Question

1.Consider the following hypotheses: H0: μ = 7,300 HA: μ ≠ 7,300 The population is normally distributed with a population standard deviation of 700. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level. (You may find it useful to reference the appropriate table: z table or t table) (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round "test statistic" values to 2 decimal places and "p-value" to 4 decimal places.)

d.	x−x− = 7,110; n = 33 Test Statistic? P-value?

2. It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportation researcher wants to determine if the statement made in the advertisement is false. She randomly test drives 34 small cars at 65 miles per hour and records the braking distance. The sample average braking distance is computed as 116 feet. Assume that the population standard deviation is 22 feet. (You may find it useful to reference the appropriate table: z table or t table)

b. Calculate the value of the test statistic and the p-value. (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.)
  

Answer 1

Solution :-

Given that ,

= 7300  

= 7110

= 700

n = 33

The null and alternative hypothesis is ,

H0 :   = 7300

Ha :    7300

This is the two tailed test .

Test statistic = z

= ( - ) / / n

= ( 7110 - 7300 ) / 700 / 33

= -1.56

The test statistic = -1.56

P- value = 2 * P ( Z < -1.56 )

= 2 * 0.0594

= 0.1188

P-value = 0.1188

( 2 )

= 120

= 116

= 22

n = 34

The null and alternative hypothesis is ,

H0 :   = 120

Ha :    120

This is the two tailed test .

Test statistic = z

= ( - ) / / n

= ( 116 - 120 ) / 22 / 34

= -1.06

The test statistic = -1.06

P - value = 2 * P ( Z < -1.06 )

= 2 * 0.1446

= 0.2892

P-value = 0.2892