Question

.. Practice: Complete problems #35, 37, and 39 on p. 392. You do NOT need to do step 5 for any of these, as there is no context for the problems. Your solutions should clearly show Steps 1-4 (and notice that Step 1 is done for you in each problem.) SKILL BUILDER 1 In Exercises 6.135 to 6.140, use the distribution and the sample results to complete the test of the hypotheses. Use a 5% significance level. Assume the results come from a random sample, and if the sam- ple size is small, assume the underlying distribution is relatively normal.no 6.135 Test Ho: u = 15 vs Ha: je > 15 using the sample results = 17.2, s = 6.4, with n = 40. 6.136 Test Hoiu = 100 vs Harui < 100 using the sample results x = 91.7, s = 12.5, with n = 30. 6.137 Test Ho:u= 120 vs Hau < 120 using the sample results x = 112.3, s = 18.4, with n = 100. 6.138 Test Ho:u= 10 vs Hiu > 10 using the sample results = 13.2, s = 8.7, with n = 12. 6.139 Test Ho:u= 4 vs Hau + 4 using the sam- ple results x = 4.8, s = 2.3, with n = 15. 6.140 Test Ho:u=500 vs Hau 500 using the sample results i = 432, s = 118, with n = 75.

Answer 1

Prob #35

Hypothesis :

$H_0:\mu =15$

$H_1:\mu >15$

LT = 17.2

S = 6.4

n=40

Z score formula

$Z=\frac{\bar{x}-\mu }{\frac{S}{\sqrt{n}}}=\frac{17.2-15 }{\frac{6.4}{\sqrt{40}}} =2.47$

P value : 0.0068 ( Note : H1 has > sign means right tailed (>)test . 1-0.9932 = 0.0068 )

Significance level = a = 0.05    (5%)

P value < 0.05 hence we reject Null hypothesis (H0) at 5% level of significance.

We have evidence or we support the claim that average is greater than 15.

Prob #37

Hypothesis :

$H_0:\mu =120$

$H_1:\mu < 120$

$\bar{x}=112.3$

S = 18.4

n = 100

Z score formula

$Z=\frac{\bar{x}-\mu }{\frac{S}{\sqrt{n}}}=\frac{112.3-120 }{\frac{18.4}{\sqrt{100}}} =-4.18$

P value : 0.0000

( Note : H1 has < sign means left tailed test test check Z table for -4.18 we get 0.0000 below -3.8 value we consider 0.0000 )

Significance level = a = 0.05    (5%)

P value < 0.05 hence we reject Null hypothesis (H0) at 5% level of significance.

We have evidence or we support the claim that average is less than 120.

Prob #39

Hypothesis :

$H_0:\mu =4$

$H_1:\mu \neq 4$

$\bar{x}=4.8$

S = 2.3

n = 15

Z score formula

$Z=\frac{\bar{x}-\mu }{\frac{S}{\sqrt{n}}}=\frac{4.8 - 4}{\frac{2.3}{\sqrt{15}}} =1.35$

P value : 0.1770 ( Note : H1 has not equal sign means two tailed test Z table gives 0.9115 = 2*(1-0.9115) =0.1770)

Significance level = a = 0.05    (5%)

P value > 0.05 hence we failse to reject Null hypothesis (H0) at 5% level of significance.

We have evidence or we support the claim that average is 4.