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

Question

A recent national survey found that high school students watched an average (mean) of 7.2 movies per month with a population standard deviation of 0.9. The distribution of number of movies watched per month follows the normal distribution. A random sample of 35 college students revealed that the mean number of movies watched last month was 6.2. 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 < 7.2B. H0: μ = 7.2; H1: μ ≠ 7.2C. H0: μ > 7.2; H1: μ = 7.2D. H0: μ ≤ 7.2; H1: μ > 7.22. State the decision rule.A. Reject H1 if z < –1.645B. Reject H0 if z > –1.645C. Reject H1 if z > –1.645D. Reject H0 if z < –1.6453. Compute the value of the test statistic.4. What is the p-value?

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Alice 1 month 2021-10-22T05:21:08+00:00 1 Answer 0 views 0

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    2021-10-22T05:22:21+00:00

    Answer:

    1) A. H0: μ ≥ 7.2; H1: μ < 7.2

    2) D. Reject H0 if z < –1.645

    3) t=\frac{6.2-7.2}{\frac{0.9}{\sqrt{35}}}=-6.573  

    4) p_v =P(z<-6.573)=2,47x10^{-11}  

    Step-by-step explanation:

    Information provided  

    \bar X=6.2 represent the sample mean for the number of movies watched last month

    \sigma=0.9 represent the population deviation

    n=35 sample size selected

    \mu_o =7.2 represent the value that we want to test    

    \alpha represent the significance level for the hypothesis test.    

    t would represent the statistic

    p_v represent the p value for the test

    1) System of hypothesis

    We want to check if college students watch fewer movies a month than high school students, and the best system of hypothesis are:

    Null hypothesis:\mu \geq 7.2      

    Alternative hypothesis:\mu < 7.2      

    A. H0: μ ≥ 7.2; H1: μ < 7.2

    2) Decision rule

    For this case we are ocnduting a left tailed test so then we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the left and we got:

    z_{crit}= -1.645

    And the rejection zone would be:

    D. Reject H0 if z < –1.645

    3) Statistic

    Since we know the population deviation the statistic is given by:

    t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

    Replacing we got:

    t=\frac{6.2-7.2}{\frac{0.9}{\sqrt{35}}}=-6.573      

    4) P value

    We have a left tailed test then the p value would be:      

    p_v =P(z<-6.573)=2,47x10^{-11}  

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