Question

A weight-loss company wants to make sure that it's clients lose more weight, on average, than they would without the company's help. An independent researcher collects data on the amount of weight lost in one month from 45 of the company's clients and finds a mean weight loss of 12 pounds. The population standard deviation for the company's clients is known to be 7 pounds per month. Data from 50 dieters not using the company's services reported a mean weight loss of 10 pounds in one month. The population standard deviation for dieters not using the company's services is known to be 6 pounds per month. Test the company's claim that using it's services results in a greater mean weight loss at the 0.05 level of significance.

Answer #1

H0: 1 <= 2

Ha: 1 > 2

Test statistics

z = (1 - 2) / sqrt [ 1 / n1 +2 / n2 ]

= (12 - 10) / sqrt [ 7^{2} / 45 + 6^{2} / 50
]

= 1.49

This is test statistics value.

p-value = P( Z > z) +

= P( Z > 1.49)

= 0.0681

Since p-value < 0.05 level, we fail to reject the null hypothesis.

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