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natta225 [31]
3 years ago
6

A recent national survey found that high school students watched an average (mean) of 7.2 movies per month with a population sta

ndard deviation of 1.0. The distribution of number of movies watched per month follows the normal distribution. A random sample of 49 college students revealed that the mean number of movies watched last month was 6.5. At the 0.05 significance level, can we conclude that college students watch fewer movies a month than high school students?
1. State the null hypothesis and the alternate hypothesis. A. H0: μ ≥ 7.2; H1: μ < 7.2
B. H0: μ = 7.2; H1: μ ≠ 7.2
C. H0: μ > 7.2; H1: μ = 7.2
D. H0: μ ≤ 7.2; H1: μ > 7.2
2. State the decision rule.
A. Reject H1 if z < –1.645
B. Reject H0 if z > –1.645
C.Reject H1 if z > –1.645
D. Reject H0 if z < –1.645
3. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
Value of the test statistic______________?
4. What is the p-value? (Round your answer to 4 decimal places.) ________________?
Mathematics
1 answer:
guajiro [1.7K]3 years ago
6 0

Answer:

Step-by-step explanation:

Download docx
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A 99% confidence interval for the difference between two proportions was estimated at 0.11, 0.39. Based on this, we can conclude
sladkih [1.3K]

Answer:

\hat p_1 -\hat p_2 \pm z_{\alpha/2} SE

And for this case the confidence interval is given by:

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Step-by-step explanation:

Let p1 and p2 the population proportions of interest and let \hat p_1 and \hat p_2 the estimators for the proportions we know that the confidence interval for the difference of proportions is given by this formula:

\hat p_1 -\hat p_2 \pm z_{\alpha/2} SE

And for this case the confidence interval is given by:

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sertanlavr [38]

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{k = 3}}}}}

Step-by-step explanation:

\sf{4.5 + 1.5k = 18 - 3k}

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Collect like terms

\longrightarrow{ \sf{4.5k = 18 -  4.5}}

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\longrightarrow{ \sf{ \frac{4.5k}{4.5}  =  \frac{13.5}{4.5} }}

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\longrightarrow{ \sf{k = 3}}

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