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3. -16 points ASWSBE13 9.E.011. My Notes + Ask Your Teacher You may need to use...

3. -16 points ASWSBE13 9.E.011. My Notes + Ask Your Teacher You may need to use the appropriate appendix table or technology

3. -16 points ASWSBE13 9.E.011. My Notes + Ask Your Teacher You may need to use the appropriate appendix table or technology to answer this question. Consider the following hypothesis test. Hoiul = 15 H: 15 A sample of 50 provided a sample mean of 14.08. The population standard deviation is 3. (a) Find the value of the test statistic. (Round your answer to two decimal places.) (b) Find the p-value. (Round your answer to four decimal places.) p-value = (c) At a = 0.05, state your conclusion. Reject Ho. There is sufficient evidence to conclude that u + 15. Reject Ho. There is insufficient evidence to conclude that u # 15. Do not reject Ho. There is sufficient evidence to conclude that # 15. Do not reject Ho. There is insufficient evidence to conclude that u 15. (d) State the critical values for the rejection rule. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused tail.) test statistics test statistic 2 State your conclusion. Reject Ho. There is sufficient evidence to conclude that u 15. Reject Ho. There is insufficient evidence to conclude that u 15. Do not reject Ho. There is sufficient evidence to conclude that 15. Do not reject Ho. There is insufficient evidence to conclude that # 15.
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Answer #1

Solution :

\mu= 15

\bar x =14.08

\sigma =3

n = 50

This is the two tailed test .

The null and alternative hypothesis is ,

H0 :  \mu  = 15

Ha : \mu    \neq 15

a ) Test statistic = z

= (\bar x - \mu ) / \sigma / \sqrt n

= (14.08 - 15 ) / 3 / \sqrt 50

= −2.168

Test statistic = z =  − 2.17

b ) P-value =0.0301

c ) \alpha = 0.05  

P-value < \alpha

0.0301 < 0.05

Reject the null hypothesis .

There is sufficient evidence to conclude that  \neq 15

d ) It is observed that ∣z∣ = 2.168 > zc​=1.96,

There is sufficient evidence to conclude that  \neq 15

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