Question

7. A study conducted a few years ago claims that the adult males spend an average of 11 hours a week watching sports on television. A recent sample of 100 adult males showed that the mean time they spend per week watching sports on television is 9.50 hours with a standard deviation of 2.2 hours.

a. Use the result to give a 95% confidence interval for the mean time they spend per week watching sports on television.

a. Does the study give good evidence that all adult males spend less than 11 hours per week watching sports on television? (Hint: apply the 4-step process of hypothesis testing)

please show work

Answer 1

ANSWER:

From the given data ,we have

Solution :

Given that,

Point estimate = sample mean = \bar x = 9.50

Population standard deviation = \sigma = 2.2

Sample size = n =100

At 95% confidence level

\alpha = 1 - 95%

\alpha = 1 - 0.95 = 0.05

\alpha/2 = 0.025

Z\alpha/2 = Z0.025 = 1.96

Margin of error = E = Z\alpha/2 * ( \sigma /\sqrtn)

= 1.96 * ( 2.2 / \sqrt100 )

= 0.43

At 95% confidence interval estimate of the population mean is,

\bar x - E < \mu < \bar x + E

9.50 - 0.43 < \mu < 9.50 + 0.43

9.07 < \mu < 9.93

( 9.07 , 9.93 )

At 95% confidence interval estimate of the population mean is : - ( 9.07 , 9.93 )

( b )

Given that ,

\mu = 11

T = 9.50

\sigma = 2.2

n = 100

The null and alternative hypothesis is ,

H0 : \mu = 11

Ha : \mu < 11

This is the left tailed test .

Test statistic = z

= (T - \mu ) / \sigma / \sqrt n

= ( 9.50 - 11 ) / 2.2 / \sqrt 100

= -6.82

The test statistic = -6.82

P- value = P (Z < -6.82 )

= 0.0000

P-value = 0.0000

\alpha = 0.05

0.0000 < 0.05

P-value < \alpha

Reject the null hypothesis .

There is sufficient evidence to the test claim .