Question

A simple random sample of size ne 15 is drawn from a population that is normal distributed. The sample mean is found to be 32.3 and the sample standard deviation is found to be 63. Determine the population means offers from 26th 0.01 level of significance Complete parts through (d) below. (*) Determine the ruland tomative hypotheses 7 26 Ho He Cote the Pue Round to tredecimal places as needed) c) the conclusion for the test OA Reject because the value is greater than the 001 level of significance OB. Reject He because the value is less than the 0.01 level of significance OC. Do not reject H, because the pain greater than the level of wignificance OD. Do not reject H, because the value is less than the 0.01 level of significance State the conclusion in context of the problem There sufficient evidence the 0.01 level of significance le conclude that the population mean is different from 26

Answer 1

a) As we are testing here whether the population mean is different from 26, therefore the null and the alternative hypothesis here are given as:

$H_0: \mu = 26$

$H_1: \mu \neq 26$

b) The test statistic here is computed as:

$t^*= \frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{32.3 - 26}{\frac{6.3}{\sqrt{15}}} = 3.87$

For n - 1 = 14 degrees of freedom, we have the p-value from t distribution tables as:
p = P(t₁₅ > 3.87) = 2*0.0008 = 0.0016

Therefore 0.002 is the required p-value here.

c) As the p-value here is 0.002 < 0.01, which is the level of significance, therefore the test is significant here.

Therefore we Reject H0 here, because p-value is less than the level of significance here.

d) As we rejected the null hypothesis in the above part, we have sufficient evidence here at 1% level of significance, to conclude that the population mean is different from 26 here.