Question

In order to conduct a hypothesis test for the population mean, a random sample of 28 observations is drawn from a normally distributed population. The resulting sample mean and sample standard deviation are calculated as 17.9 and 1.5, respectively. (You may find it useful to reference the appropriate table: z table or t table) H0 : μ 17.5 against HA: μ > 17.5 a-1. Calculate the value of the test statistic. (Round all intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.) Test statistic a-2. Find the p-value. Op-value<0.01 0.025 s p-value<0.05 0.01 s p-value<0.025 0.05 s p-value<0.10 value20.10

Answer 1

a1. The test statistics for the hypothesis test is calculated as

$T=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}}=\frac{17.9-17.5}{\frac{1.5}{\sqrt{28}}}=1.4111$

a2.The p value calculated is associated with t value and Degree of freedom 28-1=27

0.05 <P-value<0.1

a3. At the 1 % level of sognificanceConclusion:

We fail to reject since P-value is greater than 0.01

a4. Conclusion:

We fail to reject hence

We cannot conclude that mean is greater than 17.5

b1. For this given Hypothesis test

Test statistics Would be the same as

$T=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}}=\frac{17.9-17.5}{\frac{1.5}{\sqrt{28}}}=1.4111$

b2. The p-value, however, changes to for a two-tailed test as

$P value \geq 0.1$

b3. At 0.01 level of significance conclusion:

Do not reject Ho since P-value is greater than the significance level.

b4. Conclusion:

We can not conclude that the population mean differs from 17.5