Question

The owners of Johnny Jim's claim that their stores average $875,000 in annual sales. You used this information in deciding to open a store. Your store however has not come even close to these annual sales. You want to prove that the figures given to you were misleading, and that in fact the true population sales figures must be below $875,000 You collect annual sales figures from 55 randomly selected stores. The average in your sample turns out to be $880,000, with a standard deviation of $26,000. What is the relevant point estimate? O 2.242 O 55 O 26,000 O 880,000 O 875,000 What is the value of the test statistic? O 1.426:2 O-1.4262 O 0.1923 O 10.5769 O 2.2421 This test statistic has: Q A normal distribution O A t distribution with 53 degrees of freedom O A t distribution with 55 degrees of freedom Q A standard normal distribution O A t distribution with 54 degrees of freedom What is the p-value for this test? 0.9202 O 0.4601 O 0.8404 O 0.0531 O 0.0798 Session Timeout

Answer 1

Relevant point estimate in defined In Statistics as the highest probability point of inference where the best possible result or mean of a Statistical data can be drawn. In the given data here, the average of 55 stores is $880,000 and has a standard deviation of 26,000 . This means that on an average , the members of the data deviate by 26,000. The relevant point estimate here in any case would be the same as the average of the collected data.

                              i.e. the answer would be 880,000.
                

                       The value of test statistics means, by how much is the difference between the original given data and the data estimated in the statistical test performed. Here, the original data or the average from the stores collection was given as $875,000 , and after the test was done 55 random samples , the average (mean) was found out to be $880,000 . Therefore here, the value of Test statistics would be :

                              If 875,000 is = 100

                         Then 1               = 100/ 875,000

                         & 880,000        = (100/875,000) x 880,000

                                                   = 101.4262

                     Therefore , the value of Test statistics is ( 101.4262 – 100 ) = 1.4262

The degree of freedom in a test statistics is determined as the (total number of test subjects – 1).

There in this case, the degree of freedom would be (55-1) = 54.

i.e. A t distribution with 54 degree of freedom.

P-value is defined as the probability of finding the most extreme situations of the deviations in a data result. In this given case , the data of a test performed is 880,000 and the standard deviation 26,000. Now, the P-value for this test would be approximately 0.0531. This is because , the standard deviation is the chances of times that the value differs from the already given data of 875,000 . Since the deviation average is 26,000 there is almost equal chance that the nearly 50% of the members of this data would have deviated. Therefore the correct answer is 0.0531

         

Since there is 5% chance of Type I error , we will Not reject the null hypothesis , and conclude that the numbers were not misleading.

If we fond out that one of our friends , who is in management of a similar franchise, finds out that the count on the standard deviation of sales is $ 8,000 , then we would Calculate a fresh t statistic again with the new piece of information now.