Calculation Of Confidence Intervals And Hypothesis Testing

Population Mean

Sample mean,  = 45; sample standard deviation, s = 10; sample size, n = 70

1. Level of significance, α= 1 – confidence level = 1 – 0.99 = 0.01.

The formula for the confidence interval of population mean is defined as,

Where,   =  = 1.1952

z = 2.58 (obtained from the z table for α/2 = 0.005)

Margin of error =   = 3.0837

Thus, lower limit of the confidence interval = 45 – 3.0837 = 41.9163 and the upper limit = 45 + 3.0837 = 48.0837.

Therefore, 99% C.I. is (41.9163, 48.0837).

1. Here, the estimate of the required population proportion is =  =  = 0.514.

The formula for the confidence interval of population proportion estimate is defined as,

The level of significance = α = 1- 95 = 0.05.

The value of critical z at α/2 = 0.05/2 = 0.025 is obtained from the z- table as 1.96.

Standard error of =  = 0.059737.

Margin of error = 1.96  0.059737 ≈ 0.117085.

The lower limit of the 95% confidence interval =  – 0.117085 = 0.3972 and the upper limit of the confidence interval =.

Therefore, 95% C.I. is (0.3972, 0.63137)

Sample mean sample size, n = 70, sample standard deviation, s = 9.22.

1. Here, the population standard deviation is not given, so t-distribution should be used to evaluate the confidence interval. The significance level = α= 1- 95 = 05. From the t-table, it can be obtained that, t_{(α/2, n-1)} = t_(0.025,70-1) = 1.9949.

The formula of 95% confidence interval for the population mean is given by,

    t_{(α/2, n-1)}  = 21.34 ( = 21.34  

    = 21.34

Thus, the lower limit of the confidence interval = 21.34 – 2.198859 = 19.14114 and the upper limit = 21.34 + 2.198859 = 23.53886.

Therefore, the confidence interval of mean amount spent in the pet supply store is (19.14114, 23.53886).

1. The sample proportion of the customers who own a cat = p  

  =  =  =  = 0.371429.

The significance level, α = 1 – 90 = 0.10.

From the standard normal table, Z_α/2 = Z_0.10/2 = Z_0.05 = 1.645. The formula of confidence interval is

 Z_0.05   = 1.645  = 1.645  = 0.095002

The confidence interval is 0.371429  0.095002 that is, (0.276427, 0.46643).

1. In this part, the given estimated standard deviation, σ= $10; margin of error, e = $1.50; level of significance, α = 0.05.

From the z-table, the confidence coefficient Z_α/2 = Z_0.025 = 1.96.

The sample size (n) is calculated as n =  =                                                                                         =    = 170.7378.

Therefore, the sample size is 170.

1. To evaluate 90% confidence interval of the population proportion of customers, the sample size needs to be calculated.

The given information are proportion of population, π = 0.5, margin of error, e = 0.045. The level of significance, α = 1-0.90 = 0.10.

From the standard normal table, Z_α/2 = Z_0.10/2 = 1.645.

Thus, the required sample size, n =  =  =    = 334.0772.

Therefore, the sample size is 335.

1. From the calculated sample sizes in (c) and (d), the manager should take the sample size of 335 as the sample size is larger in the latter case.

The given information are sample mean,  = 5.5, standard deviation, s = 0.099, sample size, n = 50. The standard error, e = s/ = 0.099/50 = 0.014.

1. At 99% confidence level, the level of significance, α= 1-0.99 = 0.01.

The critical value of z at α = 0.01 is 2.58 (from the z-table).

Population Proportion

The confidence interval is given by 2.58.

The margin of error = 2.58 = 0.0362.

Therefore, the 99% confidence interval for the population mean weight is given by 5.5  that is, (5.4638,5.5362).

1. The problem requires to perform a t-test to check whether the mean amount of tea in the bag is 5.5 grams or not. Thus, the null hypothesis is H₀: µ= 5.5 against the alternative hypothesis H₁: µ≠ 5.5.

The sample mean is calculated from the given data of 50 observation and it is 5.497.

It is clearly seen that the sample mean lies within the confidence interval. Thus, the null hypothesis is accepted and it can be concluded that the company is meeting the requirement of containing 5.5 grams tea in the tea bag.

1. Yes, the assumptions are valid to calculate the 99% confidence interval.

1. The null and the alternative hypotheses are stated respectively as,

H₀: The proportion of population of web page visitors preferring the new design is 0.50.

H₁: The proportion of population of web page visitors preferring the new design is not 0.50

1. The type I error occurs if the null hypothesis, being true, is rejected by the investigator. The type II error occurs when the false null hypothesis is accepted by the investigator.

In the given course of problem, when the population proportion of web page visitors preferring the new design is truly 0.50 but the investigator reject the null hypothesis, then type I error occurs. Again, if the investigator accepts the null hypothesis when the population proportion of web page visitors preferring the new design is actually not 0.50, then the type II error occurs.

1. At 5% level of significance,the p-value of 0.20 is greater than α = 0.05. Thus, the null hypothesis is accepted here as there is not enough evidence to state that the null hypothesis is wrong. Therefore, the population proportion is 0.50. If the null hypothesis is rejected here then the type I error will occur.
2. If the value of α is raised, then the possibility of rejecting the null hypothesis will be increased. Thus, it will actually help to increase the probability of accepting H
3. The sample size can also be increased instead of increasing the level of significance, which will help to represent the population in better way. The larger the sample size, the better will be the representation of the population.
4. If the p-value is 0.12, then also the level of significance will be less than p-value which will again lead to accept the null hypothesis.

If the p-value is reduced to 0.06, then also the null hypothesis will not be rejected as the p-value is still greater than the significance level.

1. The sample standard deviation is = 3.99.

The null hypothesis for the required test is given by

H₀: µ  100 against the alternative hypothesis H₁: µ < 100 where µ denotes the population mean.

The test statistic, t_cal = () = -1.58 with degrees of freedom 75-1 = 74 and the significance level = 0.05

The critical value of t-statistic is -1.6657 which is greater than 0.05. Thus, the null hypothesis is accepted and there is not enough evidence to state that the mean reimbursement is less than $100.

1. The requires test of hypotheses are

H₀: π  0.10 against H₁: π > 0.10; where π is the population proportion.

Sample proportion p = 12/75 = 0.16

The calculated value of the test-statistic of this one-tail z-test =  = 1.73.

The null hypothesis is rejected as 1.73> 1.64, where 1.64 is the critical z-value at 5% level of significance.

1. The assumption behind the t-test is that the sample is drawn from normally distributed population.
2. If the sample mean is $90 then the revised value of test statistic = () = -2.51.

Clearly, -2.51< -1.64, the null hypothesis is rejected at 5% level of significance and the mean reimbursement value will be less than $100.

1. For 15 office visits the modified value of proportion = 15/75 = 0.20.
   The revised z-statistic value = = 2.89.

At 5% significance level, the H₀ will be rejected here as 2.89>1.64. Thus, there is enough proof to have the population proportion more than 0.10.

The required calculations are shown in the following table

Measurements	Gender (Female)	Gender (Male)
	18	10
	20	5.5
	10	6
	18	14
	8	10
	25	6
	22.5	12
	30	21
	20	15
	24	12
	15	10
	15	8.5
	10	25
	14	10
	15	20
	12	8
	11	8
	12	12
	15	6
	18	6
		7
		10
		10
		9
		10
		30
		10
		10
		5
		5
		10
		11
		15
		15
		7
		8
		12
		10
Count	20	38
Mean	16.625	11.0263158
Standard Deviation (SD)	5.71442082	5.40638682

The population of females is represented by population 1 and that of males is represented by population 2.

Sample Size Calculation

Thus, n₁ = 20, n₂ = 38,  = 16.625,  = 11.0263, s₁ = 5.7144, s₂ = 5.4064.

1. The hypotheses of test are given by,

H₀:  =  against H₁:  ≠  ; where   and  represents the population variances of population 1 and population 2 respectively.

The test statistics is given by,

F =  = 1.117.

Under null hypothesis, the test statistics follow F_{20-1, 38-1} = F_{19, 37} at 5% level of significance.

The p-value is given by 0.7492 which shows that the p-values is greater than 0.05 and therefore the null hypothesis is accepted. Thus, it can be concluded that the population variances are equal for both the populations.

1. It is seen that the population variances are equal. Hence, the pooled sample t test will be used.
2. The test of hypotheses are written as H₀: µ₁= µ₂ against H₁: µ₁ ≠ µ₂ where µ₁ and µ₂ are respectively the population means of population 1 and population 2.

The pooled standard deviation is calculate3d as,

S_p =  = 5.5128.

And t-statistics is calculated as t =  = 3.676 with degrees of freedom = 20+38-2 = 56.

The present test is two-tailed and the p-value is 0.0005 which shows that the p-value is less than the significance level 0.05. Therefore, the null hypothesis is rejected and finally the mean time for study of the two populations are not equal.

1. From the above statistical tests, it can be summarized that though the variances of the study time id equal for both the populations, the mean study time are different. It shows that the two populations are not independently distributed although their measures of locations are different.

The populations of Pinterest shoppers and the Facebook shoppers be respectively denoted as population 1 and population 2.

Given that, n₁ = 500, n₂ = 500,  = 153,  = 85, s₁ = 150, s₂ = 80.

1. The required test of hypotheses are H₀: =  against H₁:  ≠  ; where   and  represents the population variances of population 1 and population 2 respectively and level of significance α = 0.05.

The test statistics is given by,

F =  =  = 3.5156.

At 5% level of significance, the critical value of F with (499,499) degrees of freedom is 1.16 (obtained from the F-table).

Thus, the observed value of F is greater than the critical value of F which leads to reject the null hypothesis. Therefore there is difference in the variance of the order values of Pinterest shoppers and the Facebook shoppers.

1. The test of hypotheses are H₀: µ₁= µ₂ against H₁: µ₁ ≠ µ₂ where µ₁ and µ₂ are respectively the population means of population 1 and population 2.

The pooled variance is S² =  = 12526.15

The test statistic is t =  = 9.61.

The critical value of t statistic with df = 500+500-2 = 998 at 5% level of significance is given as 1.96.

The critical value is less than the calculated value of t-statistic. Thus, the null hypothesis is rejected and finally the mean time for study of the two populations are not equal.

1. The 95% confidence interval for the difference of mean values is obtained by the following calculation

µ₁ – µ₂ =  –  ±   = (153-85) ± 1.96  

    Thus, the 95% confidence interval is given by (53.09, 82.90).

1. Let the X variable represent the values of Twitter activity and the variable Y represent the values of Receipts.

Therefore, the simple linear regression equation is given by:

Where is the estimated value of Y corresponding to the values of X,  is the y-intercept of the sample and  is the sample slope. The values of i run from 1 to 7.

The Data Analysis Tool Pak of MS Excel is used to calculate the simple linear regression line.

t-Statistics

First, the data is written in the Excel sheet. Then, click on the “Data” tab à  Click on the “Data Analysis” tab

à Choose regression à select thee option and the data region as shown in the image belowàclick OK.

The output is obtained as,

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	6808.104703	854.9682277	7.962991468	0.00050374	4610.338906	9005.8705	4610.3389	9005.8705
Twitter Activity	0.050271541	0.003527041	14.25317587	3.0627E-05	0.041204992	0.0593381	0.041205	0.0593381

Thus, intercept b₀ = 6808.1047 and the slope b₁ = 0.0502715.

1. The y-intercept b₀ interprets that the estimate of Y will be 6808.1047 if the value of X is zero. The slope b₁indicates that for one unit increase in the value of X variable, the Y variable will be increased by approximately 0.0503 units. In other words, if the Twitter activity increases one unit, then the Receipts value will be increased by 0.0503 units.
2. The prediction of mean receipts is given as,

Therefore, the predicted value of the receipts fir a movie that ha TWITTER ACTIVITY OF 100,000 is given as  = $11835.259.

1. The model is constructed on the basis of the sample values and in the sample, it can be seen that none of twitter activity has value of 100,000. Thus, it is not justified t use this model for predicting the receipts for a movie that has Twitter activity value of 100,000.
2. The coefficient of determination is obtained in the summary output section in the Excel sheet which is shown below in the tabular format-

Regression Statistics
Multiple R	0.987916601
R Square	0.97597921
Adjusted R Square	0.971175052
Standard Error	1830.236093
Observations	7

From the table, it is seen that, the value of r² ≈ 0.9760. This value interprets that, 97.6% variation in the Receipts value is due to the variation in the Twitter activity. The remaining 2.4% variation is due to other factors which are not controlled by the Twitter activity.

1. The residual analysis is given in the following table as,

RESIDUAL OUTPUT

Twitter Activity	Receipts	Residuals
219509	14763	-3080.1604
6405	5796	-1334.0939
165128	15829	719.65628
579288	36871	941.19487
6564	8995	1856.9129
11104	7477	110.680106
9152	8054	785.810154

The residual plot is given as,

From the above graph, it is seen that there is no significant pattern in the residual plot.

1. To check whether there is any evidence of linear relationship between the two variables or not, the t-test is performed. The null and the alternative hypotheses are as follows:

H₀: β₁ = 0 (There is no linear relationship)

H₁: β₁≠ 0 (There is a linear relationship)

The ANOVA table is obtained in the Excel sheet which is given below:

ANOVA
	df	SS	MS	F	Significance F
Regression	1	680514712.65	680514712.65	203.15	0.00
Residual	5	16748820.78	3349764.16
Total	6	697263533.4

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	6808.104703	854.9682277	7.962991468	0.00050374	4610.338906	9005.8705	4610.3389	9005.8705
Twitter Activity	0.050271541	0.003527041	14.25317587	3.0627E-05	0.041204992	0.0593381	0.041205	0.0593381

From the ANOVA table, it is seen that, the value of t-statistic = 14.2532. Besides, the critical value of t-statistics at 5% level of significance is given as 2.5706 for 7-2 = 5 degrees of freedom.

Clearly, the calculated value of t –statistic is greater than the critical value which implies that the null hypothesis is rejected. Hence, there is significant linear relationship between Twitter activity and Receipts.

1. The 95% confidence interval of the mean response for a given X is defined as,

Where 11835.259,  2.5706, S = 1830.2361,  = 0.1495492.

Thus, 95% confidence interval estimate of the mean receipts for a movie having a twitter activity is 11835.259  1819.422946 that is (1005.83605, 13654.68195).

Now, the 95% prediction interval of the receipts for a single movie that has a twitter activity of 100,000, is calculated using the following formula:

Therefore, the 95% confidence interval is given as 11835.259 5044.352206 that is, (6790.906794, 16879.61121).

1. From the above results from (a) to (h), it can be stated that the Twitter activity works as a useful predictor of Receipts on the first weekend of a movie opens.

1. The ‘Invoice’ is taken as the independent variable and ‘Time’ as the dependent variable. To calculate the regression coefficients b₀and b_1, the Data Analysis Tool Pak of the MS-Excel can be used and the scatter diagram is plotted. 

Hence, the y-intercept b₀= 0.4872 and slope, b₁= 0.0123. The equation of linear regression line is Y = 0.04872 + 0.0123X.

1. The interpretation is for one unit increase in the invoice value, the time value will be increased by 0.0123units.
2. If X=150 then predicted Time value, = 0.4782+ (0.0123*150) =2.3322.
3. The coefficient of determination is 0.8623 which interprets that 86.23% of variation in the Time variable is due to effect of Invoice variable.
4. The residual plot against the Invoice variable is shown below: 

The following diagram shows the residual plot against the Time variable: 

Here, normal assumption is not specified as there is no significant pattern of the data.

1. The hypotheses for checking the appropriateness of the model is

H₀: b₁ = 0 against H₁: b₁ ≠ 0

Test statistic =  ~ t_n-2 

         = 13.71

The p-value=0.00 which fails to accept the H₀. Thus, the linear regression model is significant.

1. The results in € and (f) states that for one unit increase in invoice that is if X=151 then the time value will be increased by 0.0123 units and it will be 2.3445 units.
2. The correlation coefficient is 0.928601 which interprets that there is a strong positive linear relationship between Time and Invoice and the linear model is statistically appropriate.

The number of wins is the dependent variable and the independent variable is ERA.

1. The regression equation is obtained in MS Excel.

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	168.7804	14.81386	11.39341	5.01E-12	138.4355	199.1252	138.4355	199.1252
E.R.A.(X)	-22.7194	3.814214	-5.95652	2.06E-06	-30.5325	-14.9064	-30.5325	-14.9064

F-Statistics

The intercept b₀= 168.7804, slope b₁= -22.7194. The regression equation is

Wins = 168.7804 – 22.7194 ERA.

1. The intercept value 168.7804 is a constant value which is not altered due to the positive or negative change in ERA. The one unit increase in ERA implies 22.7194 unit decrease in Wins value.
2. For X = 4.50, the predicted value of Wins, = 168.7804 – (22.7194 *4.50) = 66.5431≈
3. The coefficient of determination 0.5589 which states that 56% variability of wins is explained by the ERA.
4. The Normal probability plot is shown below;

The scattered points have less variability around the least square line and there is no outlier. Moreover, the trendline is linear.

The residual plot is shown below:

The residual values are spread equally in the above plot in the middle.

1. The validity of the model is checked by the following test.

H₀: β₁ = 0 against H₁: β₁ ≠ 0

The p-value is 0.00 at 5% level of significance which shows that the null hypothesis is rejected and there is linear relationship between the Wins and ERA.

1. The following image shows the required calculation

In the MINITAB output above, the 95% CI for the mean wins is (60.6891, 72.3967)

1. From the output above, the 95% prediction interval is (48.6336, 84.4521).
2. The 95% confidence interval for β₁ is b₁± (t_α/2 * S_b1)

b₁= -22.71944336

S_b1 = 0.1857

From the t-table, t_(0.025,28) = 2.0484.

The C.I. is obtained as -22.7194 ± (2.0484*0.1857) that is, (-23.0998,-22.339).

1. The population consists of 30teams only. However, the “population” in general may include all the teams participated in Baseball.
2. The other independent variables considered in the model are League and Runs Scored per Game.
3. The results from (a) to (k) shows that the ERA is a useful predictor on the number of wins. The linear relationship is negative and the regression line is a good fit.

1. The given regression equation gives the predicted purchase behavior value for X₁= X₂ = 2 as  = -3.888 + (1.4492) + (1.4622) – (0.1922) = 1.174.
2. If X₁= 2 and X₂ = 7 then the predicted equation is  = -3.888 + (1.4492) + (1.462) – (0.192) = 6.584.
3. The predicted Y value for X₁= 7 and X₂ = 2 is  = -3.888 + (1.449) + (1.4622) – (0.1972) = 6.519.
4. The predicted purchase behavior when X₁= 7 and X₂ = 7, is   = -3.888 + (1.4497) + (1.4627) – (0.197) = 7.179.
5. The slope of X₁when X₂ = 2 is calculated as,  = -3.888 + (1.449 X_1i) + (1.4622) – (0.19 X_1i) = -0.964 + 1.069 X

Thus, the slope of X₁ is 1.069 when X₂ is 2.

1. The regression equation for X₂ = 7 is = -3.888 + (1.449 X_1i) + (1.4627) – (0.19 X_1i) = 6.346 + 0.119 X

The slope of X₁ is 0.119.

1. The regression equation for X₁ =2 is  = -3.888 + (1.449) + (1.462 X_2i) – (0.19 X_2i) = -0.99 + 1.082 X

The slope of X₂ is 1.082.

1. The regression equation for X₁ = 7 is

  = -3.888 + (1.449) + (1.462 X_2i) – (0.19 X_2i) = 6.255+0.132 X_2i and the slope of X₂ is 0.132.

1. From parts (e) to (f), it can be observed that with the increase of X₂ variable from 2 to 7, the value of X₁ variable decreases from 1.069 to 0.119. Thus, the effect of one independent variable X₁ on the dependent variable decreases as the value of the other variable X₂
   From part (g) to (h), the slope of the X₂variable reduces from 1.082 to 0.132 as X₁ increases from 2 to 7. Hence, , the effect of one independent variable X₂ on the dependent variable decreases as the value of the other variable X₁ increases.

The predicted purchase behavior increases from 1.174 to 6.584 as X₂ increases from 2 to 7, remaining the value of X₁ at 2 for parts (a) and (b). From (a) and (c), the  increase from 1.174 to 6.519. However, in parts (b) and (d),  increases to a very small extent. This happens due to negative effect.

1. The multiple regression line can be calculated in MS Excel using the steps DataàData Analysis à 

The regression equation Wins () = -217.4739 + (4.3522129 Field Goal %) + (1.6921454 Three-Point Field Goal %)

1. For one unit increase in Field Goal % will increase Wins by 4.3522129 units keeping Three

Pint Field Goal % constant. Again, if the Field Goal % is kept fixed, then one unit increase in the Three-Point Final Goal will increase Wins by 1.6921454 units.

1. For Final Goal =45% and Three-Point Final Goal = 37%, the predicted Wins value is = -217.4739 + (4.3522129 45) + (1.6921454 37) = 40.9851.
2. To check the validity of the regression assumptions, the plots are analysed.

The errors are normally distributed and the error points are on the trendline.

The residual plot of Field Goal % is shown below that shows there is no patterns and the variances are equal. 

The residual plot of Three-point Field Goal % is displayed below that indicates the points are not equally spread and they are concentrated between 30% and 40%.

1. The test of hypothesis is H₀: β₁= β₂ = 0 against H₁: any of the one β_i is not equal to 0.

Significance level, α = 0.05. The ANOVA table is given below:-

ANOVA
	df	SS	MS	F	Significance F
Regression	2	2397.56149	1198.7807	13.273691	9.6704E-05
Residual	27	2438.43851	90.312537
Total	29	4836

The large value of F shows that much variation in Wins is explained and as F larger than the critical F, thus the null hypothesis is rejected and the regression model is valid.

1. The p-value is obtained using F.TEST function in Excel that evaluates p-value as zero. This small value of p indicates that there is significant evidence against H₀and H₀ should be rejected.
2. The coefficient of multiple determination is 0.4958 which shows that the model is not enough significant to explain the variability in the Wins variable. Higher the value of R-square, better the explanation.
3. The value of adjusted r-square is 0.4584 which is a smaller than r-square value and it shows that only 45% of variation is explained by only that independent variable which actually affects the wins variable.
4. The test of hypothesis is H₀: β_i= 0 against H₁: β_i ≠ 0 where i=1, 2 and the t-test should be performed here with degrees of freedom (30-2-1) = 27.
5. The Excel output shows, for Field Goal % variable, the p-value is less than significance level 0.05. Therefore, the null hypothesis is rejected. However, for the Three Point Field Goal %, the p-value is greater than 0.05. Thus, H₀is accepted here. Hence, the Field Goal % variable is linearly related to Wins variable.
6. Both the independent variables have positive slope that indicates that they are positively related to the predicting variable.

The regression model is fitted in MS Excel.

1. The multiple regression equation is given by,

Wins = 87.72133+ (-18.9257) ERA + (15.96262) Runs scored per game

1. The slope of ERA variable is negative that indicates the Wins variable decreases for as the value of ERA increases. The slope is positive for Runs scored per game that interprets the increase of this variable increases the value of Wins variable.
2. For ERA = 4.00 and Runs per game = 4.0, the predicted value of Wins is

 = 87.72133- (18.9257 4) + (15.96262 4) = 63.85048 – 75.7028 + 87.72133 = 75.86901 ≈ 76.

1. The residual plots are displayed below.

The above plot shows that the data is not equally spread and they are concentrated between 3.00 and 4.50. There is no pattern.

In the above residual plot, the points are again concentrated and the points are from 3.00 to 5.00.

1. The test of hypothesis is H₀: β₁= β₂ = 0 against H₁: any of the one β_i is not equal to 0.

Significance level, α = 0.05. The ANOVA table is given below:-

ANOVA
	df	SS	MS	F	Significance F
Regression	2	3828.662	1914.331	102.282	2.51392E-13
Residual	27	505.3375	18.7162
Total	29	4334

The large value of F-statistic indicates that the null hypothesis is rejected and the model is valid.

1. The p-value is obtained using F.TEST function in Excel that evaluates p-value as zero. This small value of p indicates that there is significant evidence against H₀and H₀ should be rejected.
2. The value of r-square is 0.883401587 that 88% of the variability is explained by dependent variables. The model is also a good fit.
3. The adjusted r-square is 0.8748 which is a smaller than r-square value and it shows that only 87% of variation is explained by only that independent variable which actually affects the Wins variable.
4. The test of hypothesis is H₀: β_i= 0 against H₁: β_i ≠ 0 where i=1, 2 and the t-test should be performed here with degrees of freedom (30-2-1) = 27.
5. At 5% level of significance, the p-values for both the independent variables are 0 that lead to reject the null hypothesis as p-values are less than 0.05.
6. The 95% confidence interval of the population slope of ERA variable is of the form

b₁ ± 1.96

Now,  = (0.402273567/ 2.166309458) = 0.185695338

The 95% C.I. is ((-18.9257) ± 1.96  0.185695) that is, (-19.28966, -18.56174).

1. The absolute value of the coefficient actually determines the importance of the predictor in the regression model. IN this problem, the mod value of the coefficient of ERA is larger than that of Runs scored per game. Thus, the pitching, as measured by ERA is more important in predicting wins.s