Question

Consider the following hypotheses:

H0: μ = 290
HA: μ ≠ 290

The population is normally distributed with a population standard deviation of 79. (You may find it useful to reference the appropriate table: z table or t table)

a-1. Calculate the value of the test statistic with x−x− = 303 and n = 50. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)
  


a-2. What is the conclusion at the 5% significance level?
  

Reject H0 since the p-value is less than the significance level.

Reject H0 since the p-value is greater than the significance level.

Do not reject H0 since the p-value is less than the significance level.

Do not reject H0 since the p-value is greater than the significance level.

a-3. Interpret the results at αα = 0.05.

We conclude that the population mean differs from 290.

We cannot conclude that the population mean differs from 290.

We conclude that the sample mean differs from 290.

We cannot conclude that the sample mean differs from 290.

b-1. Calculate the value of the test statistic with x−x− = 274 and n = 50. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)
  


b-2. What is the conclusion at the 1% significance level?
  

Do not reject H0 since the p-value is less than the significance level.

Do not reject H0 since the p-value is greater than the significance level.

Reject H0 since the p-value is less than the significance level.

Reject H0 since the p-value is greater than the significance level.

b-3. Interpret the results at αα = 0.01.

We cannot conclude that the population mean differs from 290.

We conclude that the population mean differs from 290.

We cannot conclude that the sample mean differs from 290.

We conclude that the sample mean differs from 290.

Answer 1

Solution :

= 290

=303

=79

n = 50

This is the two tailed test .

The null and alternative hypothesis is ,

H0 :    = 290

Ha :     290

Test statistic = z

= ( - ) / / n

= (303-290) /79 / 50

= 0.37

Test statistic = z = 0.37

P(z > 0.37 ) = 1 - P(z < 0.37 ) =1 - 0.6443

P-value = 2 * 0.3557 =0.7114

a ) = 0.05  

P-value >

0.7114 > 0.05

Do not reject H0 since the p-value is greater than the significance level.

We cannot conclude that the population mean differs from 290.

b ) = 0.01  

P-value >

0.7114 > 0.01

Do not reject H0 since the p-value is greater than the significance level.

We cannot conclude that the population mean differs from 290