Analysis Of Population Mean Weight-Loss Test Results And Relationship Between Interest Rates And Stock Prices

99% Confidence Interval for the population mean weight-loss

Part 1a)

99% Confidence Interval for the population mean weight-loss

 Part 1b)

Ho: µ = 20      

H1: µ ≠ 20 (Two tailed test)  

èt = (x-µ)/standard error

èT-statistics = 2.135372112

èT critical value (α=0.01, 2 tailed) = 2.58

At 1% significance level, T statistics < T critical value stating that the null hypothesis will be accepted that the population mean equals 20 against the two-sided alternative.

Part 1c)

Ho: µ = 20      

H1: µ < 20 (One tailed test)

èt = (x-µ)/standard error      

èT statistics = 2.135372112

èT critical value (α=0.05, one tailed) = 1.645

At 5% significance level, T statistics > T critical value stating that the null hypothesis will be rejected that the population mean weight-loss is lower than 20 pounds at the 5% level against the one-sided alternative hypothesis.

Part 1d)

P value results based on excel

=1-NORMDIST(X, Mean, Standard Deviation,TRUE)

=1-NORMDIST(20,16.74,1.699,TRUE)

èp value = 0.016365313

Part 1e)

P value results based on excel

=1-NORMDIST(X, Mean, Standard Deviation,TRUE)

=1-NORMDIST(20,16.74,1.699,TRUE)

P value = 0.016365313

The calculated p-value of .0164 is the probability of committing a Type I Error (chance of getting it wrong). A p-value of .016 is a considerable low probability of making a mistake, so it can be concluded that the averages are different and would not fall back to the null hypothesis of population mean being 20.

Part 1f)

Ho: µ = 20                  

H1: µ < 20                  

èt = (x-µ)/standard error                  

èStandard error = 1.58113883 = 5/SQRT(10)         

èt statistics = 0.632455532  

èt critical value (α=0.05, one tailed) =1.645

Hence, null hypothesis accepted.      

èCritical value

=NORMINV(Probability (1-α),mean,Std dev)

=NORMINV(1-0.05,19,1.5811)

= 21.60074194           

Beta

=NORMINV(1-0.05,19,1.5811)

=0.844326126 , this states that 84.32% failed to reject null hypothesis which is false, (false negative)

Power  = (1-β) = 1- 0.844326126 = 0.155673874     

Since the p-value of 0.2160 is so high, the null hypothesis provides a good explanation of the data. Now, based on the decision, the risk of making a type 1 error is more. It is not possible at this point to make Type 11 error. The probability of Type 2 error is only 0.1556.

Question 2

Part 2a)

Descriptive Statistics of Independent Samples

 Part 2b)

Ho:σ₁= σ₂

Ha: σ₁≠ σ₂

èy?₁=59.833, σ₁=20.390, var = 415.7879

èy?₂=50.25, σ₂ =20.045, var = 401.8409

 Part 2e)

The required condition for part c) and d) satisfied at the one tailed tests at df 22 at % significant level which makes the results directional assuming equal variances for two samples whether independent samples or matched samples.

Part 2f)

Independent sample t test for equal variances is undertaken because the two samples are taken from different population. Also, where there is equal number of data points in both then equal various test is considered whereas pairing of data is beneficial as it helps in analyzing the pre and post variation for the same data with same data points. These differences occur because matched pair is from the same population and independent t test is undertaken from two different populations.

Question 3

Part 3a)

As per the basic understanding, interest rates and stock prices have inverse relationship because the high interest rates increases the costs of the companies that in turn lower the profits followed by stock prices. However, the rising interest rate can be beneficial to reflect positive trends in the economy.

Part 3b)Yes, the scatter plot showing a downward increasing trend further justifies an inverse relationship between interest rates and sock prices.

Part 3c)

Y_t = β₁ + β₂X_t+ ε

èS&P 500 = 1229.341 + 99.4014The R square came out as 0.4249962 which further states that around 42.49% variations are explained by the independent variables in the model. The t statics is -43.83 highlighting inverse relationship between the variables but being statistically valid as the p value is less than 0.05 at 5% significance level.

Part 3d)

As per the analysis, Treasury bill (interest rates) is the independent variable for the model that affects the stock prices. Also, a change in Treasury bill would lead to 99.4014 units changes in the stock prices results inversely. Yes, the sign of the slope coefficient is consistent with the expectation of the given results assumed.

Part 3e)

Ho: The interest rates and stock prices have no significant relationship       

H1: The interest rates and stock prices have inverse significant relationship (one tailed)

èt = (x-µ)/standard error                  

èStandard error = 22.676 = 5/SQRT(28)     

èt statistics = -4.38343

èt critical value (α=0.05, one tailed) =-1.703

Hence, the null hypothesis will be rejected because the interest rates and stock prices have inverse significant relationship. 

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