A sample of 37 observations is selected from a normal population. The sample mea

Question

A sample of 37 observations is selected from a normal population. The sample mean is 25, and the population standard deviation is 5. Conduct the following test of hypothesis using the 0.05 significance level. H0 : 24 H1 : > 24 a. Is this a one- or two-tailed test? "Two-tailed"-the alternate hypothesis is different from direction. "One-tailed"-the alternate hypothesis is greater than direction. b. What is the decision rule? (Round your answer to 3 decimal places.) H0, when z > c. What is the value of the test statistic? (Round your answer to 2 decimal places.) Value of the test statistic d. What is your decision regarding H0? Do not reject Reject There is evidence to conclude that the population mean is greater than 24. e. What is the p-value? (Round your answer to 4 decimal places.) p-value

Explanation / Answer

Set Up Hypothesis
Null Hypothesis H0: U<=24
Alternate Hypothesis H1: U>24
Test Statistic
Population Mean(U)=24
Given That X(Mean)=25
Standard Deviation(S.D)=5
Number (n)=37
we use Test Statistic (Z) = x-U/(s.d/Sqrt(n))
Zo=25-24/(5/Sqrt(37)
Zo =1.2166
| Zo | =1.2166
Critical Value
The Value of |Z | at LOS 0.05% is 1.64
We got |Zo| =1.2166 & | Z | =1.64
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value : Right Tail - Ha : ( P > 1.2166 ) = 0.1119
Hence Value of P0.05 < 0.1119, Here We Do not Reject Ho

[ANSWERS]
1. ONE TAILED
2. Reject Ho when Ho>1.64
3. Zo =1.2166
4. Do n't reject Ho
5. P-value = 0.1119

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