Answer to Question #212305 in Statistics and Probability for Manasa Reddyrajula

Question #212305

Consider the following information to validate the hypothesis that the mean of the population


is at most 30 at 5% level of significance.


Sample size = 45, mean of the sample = 38, variance = 4



What will be conclusion if the level of significance is taken as 1%? Mention the observations, if


any.



1
Expert's answer
2021-07-01T09:48:47-0400

1. The following null and alternative hypotheses need to be tested:

H0:μ30H_0: \mu\leq 30

H1:μ>30H_1:\mu>30

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.


Based on the information provided, the significance level is α=0.05,\alpha=0.05,

df=n1=451=44df=n-1=45-1=44 degrees of freedom, and the critical value for a right-tailed test is tc=1.68023.t_c=1.68023.

The rejection region for this right-tailed test is R={t:t>1.68023}.R=\{t:t>1.68023\}.

The t-statistic is computed as follows:


t=xˉμs/n=38304/45=13.4164t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{38-30}{4/\sqrt{45}}=13.4164

Since it is observed that t=13.4164>1.68023=tc,t=13.4164>1.68023=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for α=0.05,df=44,t=13.4164,\alpha=0.05, df=44, t=13.4164,right-tailed is p=0,p=0, and since p=0<0.05=α,p=0<0.05=\alpha, it is concluded that the null hypothesis is rejected.


Therefore, there is not enough evidence to claim that the population mean μ\mu

μ\mu  is at most 30, at the α=0.05\alpha=0.05 significance level.



2. The following null and alternative hypotheses need to be tested:

H0:μ30H_0: \mu\leq 30

H1:μ>30H_1:\mu>30

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.


Based on the information provided, the significance level is α=0.01,\alpha=0.01,

df=n1=451=44df=n-1=45-1=44 degrees of freedom, and the critical value for a right-tailed test is tc=2.414134.t_c=2.414134.

The rejection region for this right-tailed test is R={t:t>2.414134}.R=\{t:t>2.414134\}.

The t-statistic is computed as follows:


t=xˉμs/n=38304/45=13.4164t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{38-30}{4/\sqrt{45}}=13.4164

Since it is observed that t=13.4164>2.414134=tc,t=13.4164>2.414134=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for α=0.05,df=44,t=13.4164,\alpha=0.05, df=44, t=13.4164,right-tailed is p=0,p=0, and since p=0<0.01=α,p=0<0.01=\alpha, it is concluded that the null hypothesis is rejected.


Therefore, there is not enough evidence to claim that the population mean μ\mu

μ\mu  is at most 30, at the α=0.01\alpha=0.01 significance level.


Therefore there is not enough evidence to claim that the population mean μ\mu

μ\mu  is at most 30, at α=0.05\alpha=0.05 significance level and at α=0.01\alpha=0.01 significance level.


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