Have American males (AMs) gotten heavier over the last 20 years? A random sample 77 AMs in 2019 had a mean weight of x=189.030 pounds. A random sample 93 AMs in 1999 had a mean weight of y=184.795 pounds. It is recognized that the true standard deviation of 2019 AMs weights is σx=14.04 pounds while it is recognized that the true standard deviation of 1999 AMs weights is σy=10.03 pounds. The true (unknown) mean of 2019 AMs weights is μx pounds, while the true (unknown) mean of 1999 AMs weights is μy pounds. Weights are known to be a normally distributed. In summary: TypeSample SizeSample MeanStandard Deviation2019 Data(X)77189.03014.041999 Data(Y)93184.79510.03 i) If we used this data to test H0:μx−μy=0 against the alternative Ha:μx−μy&gt;0 then what would the p value have been? j)If we used this data to test H0:μx−μy=0 against the alternative Ha:μx−μy≠0 then what would the p value have been?

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Have American males (AMs) gotten heavier over the last 20 years? A random sample 77 AMs in 2019 had a mean weight of x=189.030 pounds. A random sample 93 AMs in 1999 had a mean weight of y=184.795 pounds. It is recognized that the true standard deviation of 2019 AMs weights is  σx=14.04 pounds while it is recognized that the true standard deviation of 1999 AMs weights is  σy=10.03 pounds. The true (unknown) mean of 2019 AMs weights is  μx pounds, while the true (unknown) mean of 1999 AMs weights is  μy pounds. Weights are known to be a normally distributed. In summary:  TypeSample SizeSample MeanStandard Deviation2019 Data(X)77189.03014.041999 Data(Y)93184.79510.03  i) If we used this data to test H0:μx−μy=0 against the alternative  Ha:μx−μy&amp;gt;0 then what would the p value have been?  j)If we used this data to test H0:μx−μy=0 against the alternative  Ha:μx−μy≠0 then what would the p value have been?

Sculd1987 · Accepted Answer

Step 1   i) Null hypothesis:   H0:μx−μy=0   Alternative hypothesis:   Ha:μx−μy&amp;gt;0   The test statistic is as follows:  Z=x―1−x―2σ12n1+σ22n2  =189.030−184.795(14.04)277+(10.03)293  =2.2192  Step 2   Thus, the test statistic value is 2.2192.   Computation of P-value:   The P-value can be obtained using the excel formula “= (1-NORM.S.DIS (2.2192,1)”.   The p-value is 0.01323.   Step 3   j) Null hypothesis:   H0:μx−μy=0   Alternative hypothesis:   Ha:μx−μy≠0   Thus, the test statistic value is 2.2192.   Computation of P-value:   The P-value can be obtained using the excel formula “=2×(1−NORM.S.DIS(2.2192,1)”.   The p-value is 0.02647.

Have American males (AMs) gotten heavier over the last 20 years? A ran

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