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in a recent study of 45 ninth grade students, the mean number of hours spent studying was 6.8. the standard deviation was 1.3.
a. find the best point estimate of the mean.
b. find the 90% confidence interval.
c. find the 95% confidence interval.
d. find the 99% confidence interval.

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Ответ:

jcross11nate

03.10.2021

Mathematics

$\longrightarrow{\green{A.\:5.5 \: n - \frac{1}{2} \: p - 2}}$ ✔

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}$

$1 + 4.25 \: n + \frac{3}{2} \: p - 3 + ( - 2 \: p) + \frac{5}{4} \: n$

Combining like terms, we have

$= 4.25 \: n + \frac{5}{4} \: n + \frac{3}{2} \: p - 2 \: p + 1 - 3 \\ \\= 4.25 \: n + 1.25 \: n + 1.5 \: p - 2 \: p - 2 \\\\ = 5.5 \: n - 0.5 \: p - 2 \\ \\ = 5.5 \: n - \frac{1}{2} \: p - 2$

$\bold{ \green{ \star{ \orange{Mystique35}}}}⋆$

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Show work in a recent study of 45 ninth grade students, the mean number of hours spent studying was 6.8. the standard deviation was 1.3. a. find the best point estimate of the mean. b. find the 90% confidence interval. c. find the 95% confidence interval. d. find the 99% confidence interval.