In this example of paired data, a shoe manufacturer claims that athletes can increase their vertical
jump heights using the manufacturers training shoes. The vertical jump heights
of eight randomly selected athletes are measured. After the athletes have used
the training shoes for eight months, their vertical jump heights are measured
again. The vertical jump heights in inches for each athlete are shown in the
table. Assuming the vertical jump heights are normally distributed, is there enough
evidence to support the manufacturers claim at alpha=.10 This is the
first time we’ve had a 10% alpha. In your homework in
MyMathLab, you would get something very similar to this but you probably won’t
have this row for differences. The manufacturer is
assuming the original jump height will be less than the new jumping height. Be aware of that when you enter the information into Statcrunch. How do we state our Ho and Ha?
Our null hypothesis is that the average difference=0.
If these shoes made no impact at all, the difference in original and new jumping heights would be 0 The manufacturers claim is that the average difference in jumping height>0. Our confidence level is 90% so alpha=.10 This is a right-tail test from this>sign.
Our sample size is 8 so degrees of freedom is 7 and our critical value is 1.415 I found this on the T table. We had a
degrees of freedom of 7 ( 1 less than a sample size of 8) and this is a
one-tailed probability at .10 alpha. The 7 degrees of freedom row and the .01 column intersect at 1.415 The manufacturer is claiming that the average difference in jumping heights is going to be greater than zero. You could switch the signs so we are taking new minus original to have positive differences. Or use new jumping height minus original jumping height. Sample one is the new jumping height and sample two the original jumping height. We want my differences to be
positive to reflect the new jumping height compared to the original jumping height. Your homework will have a prompt or small rectangle beside the data. When you click the rectangle, your data will spill into Statcrunch. Click Stats, down to T Stats over and down to Paired. For this example, Sample One is
var two and Sample Two is var one (When I typed my data into StatCrunch, I placed the original jumping in var one and the new jumping heights in var two.) You can opt to save the differences and come back
at that later. The null is that the mean of the differences is=. The alternative is that the mean of the differences is greater. Click Compute. StatCrunch gives
us the mean difference of 1.75 inch increase in jumping height over an
eight-month period. Our t-stat is 2.3 with a
p-value of 0.0262 That test stat is much higher than the 1.415 critical value.The tells us this fall in the rejection zone and is outside the normal range. The p-value for our test stat is low. If the p is low, the null must go. Both methods
support the decision to Reject the Null Hypothesis. We can go to Options, down to Edit and select Confidence Interval. This was a one tailed test with alpha=10 All 10%of that alpha was in that
right tail. The two-tailed confidence interval is a lthat would have 10% alpha in each tail would be a confidence interval
at a 0.8 confidence level. Click Compute. StatCrunch gives us a confidence Interval with the lower limit at 0.688 inches to an upper limit of 2.81 inches. We are 80% confident that average increase in jumping height is going to fall in that interval.
Zero is not in that interval so it must be that the shoes do increase jumping
heights. Beware, your MyMathLab might ask for the standard deviation of the differences too. You can compute the summary stats for the difference
column. Stats, Summary Stats, With Data. Select the difference column. Click Compute. Locate the standard deviation of 2.1213. You would also need the mean and the standard deviation of the differences if you were doing this by hand. We’ll finish our problem by hand. Our average difference
was 1.75. We’re testing the average
difference of zero so entering this summary information into this formula: the average difference – 0 divided by the division of the standard deviation of the difference (2.1213) by the square root of 8 is the other to come up with our test stat of 2.33 We search to see where this falls on the row 7 of the table. This value lies between the 0.05 and 0.025 (If
you don’t use Statcrunch, you would estimate where 2.33 falls in that row of the T table. Be sure that you are finding where it falls and following those to columns up to top row with alphas in 1 tail. That ends up right here
between 0.025 and 0.05 We draw the picture and plot the critical value so we can see that the test statistic falls out in that rejection zone far to
the right of the critical value. To get exact p-value we must use Statcrunch. To calculate the confidence interval, enter the average difference plus or minus the critical value times the standard deviation of the average difference divided by the square root of the sample size. The confidence interval has a lower limit of 0.689 inches and an upper limit of 2.811 inches. 0 is not in this interval so we are 80% confident that the average difference in jumping heights is greater than 0. All three methods support the decision to Reject the Null Hypothesis.There is sufficient evidence to support the shoe manufacturers claim that the
shoes are effective at increasing and improving the jumping height of athletes.

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Dennis Veasley

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