The following hypothesis test is to be conducted. H0: Median...

The following hypothesis test is to be conducted. H0: Median ? 150 Ha: Median > 150 A sample of 31 provided 23 observations greater than 150, 3 observations equal to 150, and 5 observations less than 150. Use Table 1 of Appendix B. What is the mean and standard deviation of the normal distribution that can be used for this statistical test (to 4 decimals if necessary)? Mean 13.5 Standard deviation 2.5981 What is the value of the test statistic (to 2 decimals)? -46.72; Use ? = .01, what is the p-value (to 4 decimals)? 0.999 What is your conclusion? The population median is less than 150.

Transcript

00:01 All right, so in our given problem, we have been given that we have a null hypothesis which states that the median is less than equal to 150 and an alternate hypothesis which states that the median is greater than 150.

00:13 And we're supposed to conduct a hypothesis test but interestingly we have been given the number of samples for each of the observations.

00:20 So for a total of 31 observations, 23 plus 3 plus 5 is 31, it's found out that 23 of them are greater than 150, 5 are less than 150, and 3 are exactly equal to 150.

00:36 But this problem is a little complicated because we have to conduct a hypothesis test and that seems slightly tricky out here at this point.

00:46 So what we'll firstly do is that we have to define a null hypothesis and alternate hypothesis in terms of a probability.

00:58 So if we define it in terms of a probability, what this essentially states is that the, um, i'm sorry, one second, let me just find a better way to explain this because i don't want to complicate it too much.

01:27 Okay, so our first step would be to obviously convert this into a probability as i mentioned.

01:32 Now for each value there is a probability that it's either greater than equal to 0 .5 or it is less than equal to 0 .5.

01:48 So the null hypothesis we state is that the probability is less than equal to 0 .5 and the alternate hypothesis is that if the value is greater than 150 the chance of that happening is also 0 .5, right.

02:08 So we actually have to, um, we've been given now that there are 23 values greater than 150.

02:15 So there are 23 such samples where the probability of the sample being greater than 150 is 0 .5 and 8 such samples where the probability is less than equal to 0 .5, whether it's 3 equal to's or 5 of those are less than 150.

02:32 So we have to basically redefine our null hypothesis because what we have to do is in order to have a sign test we can either have values that are equal to 0 .5 or not equal to 0 .5, right.

02:48 So the probability has to be equal to 0 .5 or not equal to 0 .5.

02:51 So what that is telling me is that either the number is 150 or it's not 150.

03:01 Either the number is greater than 150 or it's less than 150.

03:05 So for that we'll have to neglect our samples which are equal to 150.

03:12 And so if we transform this, the null hypothesis becomes that the values that are greater than 150 or 0 .05 or they're not greater than 150 which is that they're less than 150 and that is p not equal to 0 .5.

03:39 And so the number of total samples that we take for our normal distribution would be 28 and we have a binomial in this case and a probability is 0 .5.

03:54 So from this we have to calculate the mean and the standard deviation...

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