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oee [108]
3 years ago
6

A recent national survey found that high school students watched an average (mean) of 7.6 movies per month with a population sta

ndard deviation of 0.9. The distribution of number of movies watched per month follows the normal distribution. A random sample of 39 college students revealed that the mean number of movies watched last month was 7.1. At the 0.05 significance level, can we conclude that college students watch fewer movies a month than high school students?
Mathematics
1 answer:
Maksim231197 [3]3 years ago
8 0

Answer:

We conclude that college students watch fewer movies a month than high school students at the 0.05 significance level.

Step-by-step explanation:

We are given that a recent national survey found that high school students watched an average (mean) of 7.6 movies per month with a population standard deviation of 0.9.

A random sample of 39 college students revealed that the mean number of movies watched last month was 7.1.

Let \mu = <u><em>mean number of movies watched by college students last month.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 7.6 movies      {means that college students watch higher or equal movies a month than high school students}

Alternate Hypothesis, H_A : \mu < 7.6 movies     {means that college students watch fewer movies a month than high school students}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                          T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean number of movies watched last month = 7.1

            σ = population standard deviation = 0.9

            n = sample of college students = 39

So, <u><em>the test statistics</em></u>  =  \frac{7.1-7.6}{\frac{0.9}{\sqrt{39} } }

                                       =  -3.47

The value of z test statistics is -3.47.

<u>Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.</u>

Since our test statistic is less than the critical value of z as -3.47 < -1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that college students watch fewer movies a month than high school students.

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A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40
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a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a. How large was the sample used in this survey?

We have that s = 25, \sigma = 400. We want to find n, so:

s = \frac{\sigma}{\sqrt{n}}

25 = \frac{400}{\sqrt{n}}

25\sqrt{n} = 400

\sqrt{n} = \frac{400}{25}

\sqrt{n} = 16

(\sqrt{n})^2 = 16^2[tex][tex]n = 256

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

15 is the bounds with want, 25 is the standard error. So

Z = 15/25 = 0.6 has a pvalue of 0.7257

Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

45.14% probability that the point estimate was within ±15 of the population mean

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