A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim? Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs) d = (golf score after using the newly designed golf clubs) − (golf score before using the newly designed golf clubs) . Use a significance level of α=0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs.
Score (old design)88 75 93 77 80 79 76 83
Score (new design)81 78 89 74 82 75 72 76
Step 1 of 5:State the null and alternative hypotheses for the test.
Step 2 of 5:Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.
Step 3 of 5:Compute the value of the test statistic. Round your answer to three decimal places.
Step 4 of 5:Find the p-value for the hypothesis test. Round your answer to four decimal places.
Step 5 of 5:Draw a conclusion for the hypothesis test.
Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.
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Step 2:State the null and alternative hypotheses for the test.
VIEWStep 3: The value of the standard deviation of the paired differences.
VIEWStep 4: The value of the test statistic
VIEWStep 6: Decision rule
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