Consider the following hypothesis test: H0: ? = 15 Ha: ? ?...

Consider the following hypothesis test: H0: ? = 15 Ha: ? ? 15 A sample of 50 provided a sample mean of 14.11. The population standard deviation is 3. Enter negative value as negative number. a. Compute the value of the test statistic (to 2 decimals). -2.10 b. What is the p-value (to 4 decimals)? Use the value of the test statistic rounded to 2 decimal places in your calculations. c. Using ? = 0.05, can it be concluded that the population mean is not equal to 15? d. What is the rejection rule using the critical value? Reject H0 if z is the lower critical value and is the upper critical value. Using ? = 0.05, what are the critical values for the test statistic? (+ or -) (to 2 decimals) Can it be concluded that the population mean is not equal to 15?

Transcript

00:01 Hi, i'm david and i'm here to help you answering your question.

00:03 Now let me bring up your question here.

00:06 In the question here we are going to discuss about the hybrid testing.

00:11 Now we have the known hypothesis when the m equal to the 15 versus the alternative hypothesis mean not equal to the 15.

00:20 Notice that this will be the 2 -tel test.

00:24 Now we are given the n equal to the 50, the x bar equal to the 14 points.

00:31 11 and the sigma equal to the 3 because that will be the population standard division the question asked which you find the test statistics the formula equal to the xb minus mu over sigma divided by square of n so we get the 14 .11 minus the mule will be the 15 divided by the 3 the over square root of the 50 so if we compute we get equal to dividing by 3 times square root plus 15 equal to the minus 2, i need to route to the 2 decimal places and i see you got it correctly good job for that now next i will compute the b value so b value equal to the probability of because it will be the 2 tenths so h times 2 in front and the z it could be smaller than the test statistics now to find this value on the probability and it's jumping on the z table so let me combine the z table and i put the table on the right here.

01:35 Notice at the table i bring here it has the negative g score with the probability on the left tail so it is compatible with this probability when the g score equal to the minus 2 .1 and the 0 the corresponding probability equal to the 0 .0179 then we get equal to 2 times 0 .0179 it will equal to the 0 .0358 and when we compare this value here, 0 .0358, we see it's smaller than the unfound will be the zbon 05.

02:13 So therefore we're going to reject the noon hypothesis...

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