Question

Asked Mar 4, 2019

Step 1

Given that water temperature measured on 9 randomly chosen days is normally distributed with mean 98^{o}F and standard deviation of temperature is given as 2^{o}F. We need to test if water temperature is acceptable.

Null hypothesis : mean water temperature = 100^{o}F

alternative hypothesis : mean water temperature < 100^{o}F

Here we know population standard deviation so we use a Z test. Test statistic is given by below mentioned formula . y substituting values we get test statistic = -3.

referring to Z table the Z score for probability 0.0505 is -1.64 and Z score for probability 0.4945 is -1.65. So the Z score for probability 0.95 is average of -1.64 and -1.65 = -1.645. At 0.05 significance level the critical Z score is -1.645.

Comparing to test statistic Z score is way less than critical Z score so we reject the null hypothesis.

So we can say at 0.05 significance level the water temperature is less than 100. So the water temperature is acceptable.

Step 2

The p value for Z score -3 is 0.0013 with reference to the Z table shown below.

Step 3

Given that water has true mean temperature of 104oF. we need to find the probability of accepting the null hypothesis.

we accept the nul hypothesis is p value is greater than 0.05. So the critical score has to be greater than -1.645. The temperature with respect to Z...

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