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Update _20-sa-inference-two-means.qmd #550

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4 changes: 2 additions & 2 deletions exercises/_20-sa-inference-two-means.qmd
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1. The hypotheses should use population means ($\mu$) not sample means ($\bar{x}$), the null hypothesis should set the two population means equal to each other, the alternative hypothesis should be two-tailed and use a not equal to sign.
\addtocounter{enumi}{1}

1. $H_0: \mu_{0.99} = \mu_{1}$ and $H_A: \mu_{0.99} \ne \mu_{1}.$ p-value $<$ 0.05, reject $H_0.$ The data provide convincing evidence that the difference in population averages of price per carat of 0.99 carats and 1 carat diamonds are different.
1. $H_0: \mu_{0.99} = \mu_{1}$ and $H_A: \mu_{0.99} \ne \mu_{1}.$ p-value $<$ 0.05, reject $H_0.$ The data provide convincing evidence that the difference in population averages of price per carat of 0.99 carats and 1 carat diamonds is different.
\addtocounter{enumi}{1}

1. \(a\) We are 95% confident that the population average price per carat of 0.99 carat diamonds is \$2 to \$23 lower than the population average price per carat of 1 carat diamonds. (b) We are 95% confident that the population average price per carat of 0.99 carat diamonds is \$2.91 to \$21.10 lower than the population average price per carat of 1 carat diamonds.
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1. The difference is not zero (statistically discernible), but there is no evidence that the difference is large (practically important), because the interval provides values as low as 1 lb.
\addtocounter{enumi}{1}

1. $H_0: \mu_{0.99} = \mu_{1}$ and $H_A: \mu_{0.99} \ne \mu_{1}$. Independence: Both samples are random and represent less than 10% of their respective populations. Also, we have no reason to think that the 0.99 carats are not independent of the 1 carat diamonds since they are both sampled randomly. Normality: The distributions are not extremely skewed, hence we can assume that the distribution of the average differences will be nearly normal as well. $T_{22} = -2.7$, p-value = 0.0131. Since p-value less than 0.05, reject $H_0$. The data provide convincing evidence that the difference in population averages of price per carat of 0.99 carats and 1 carat diamonds are different.
1. $H_0: \mu_{0.99} = \mu_{1}$ and $H_A: \mu_{0.99} \ne \mu_{1}$. Independence: Both samples are random and represent less than 10% of their respective populations. Also, we have no reason to think that the 0.99 carats are not independent of the 1 carat diamonds since they are both sampled randomly. Normality: The distributions are not extremely skewed, hence we can assume that the distribution of the average differences will be nearly normal as well. $T_{22} = -2.7$, p-value = 0.0131. Since p-value less than 0.05, reject $H_0$. The data provide convincing evidence that the difference in population averages of price per carat of 0.99 carats and 1 carat diamonds is different.
\addtocounter{enumi}{1}

1. We are 95% confident that the population average price per carat of 0.99 carat diamonds is \$2.96 to \$22.42 lower than the population average price per carat of 1 carat diamonds.
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