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m_a_m_a [10]
2 years ago
8

Hypothesis Testing

Mathematics
1 answer:
Yuri [45]2 years ago
8 0

Answer:

<u>Problem 1</u>: We conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

<u>Problem 2</u>: We conclude that the volume of Google stock has changed.

Step-by-step explanation:

<u>Problem 1:</u>

We are given that in a recent survey conducted by Pew Research, it was found that 156 of 295 adult Americans without a high school diploma were worried about having enough saved for retirement.

Let p = <em>proportion of adult Americans without a high school diploma who are worried about having enough saved for retirement</em>

So, Null Hypothesis, H_0 : p \leq 50%    {means that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement}

Alternate Hypothesis, H_A : p > 50%     {means that a majority of adult Americans without a high school diploma are worried about having enough saved for retirement}

This is a right-tailed test.

The test statistics that would be used here is <u>One-sample z-test</u> for proportions;

                       T.S.  =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of adult Americans who were worried about having enough saved for retirement = \frac{156}{295} = 0.53

           n = sample of adult Americans = 295

So, <u><em>the test statistics</em></u> =  \frac{0.53-0.50}{\sqrt{\frac{0.50(1-0.50)}{295} } }

                                    =  1.03

The value of z-test statistics is 1.03.

<u>Also, the P-value of the test statistics is given by;</u>

              P-value = P(Z > 1.03) = 1 - P(Z \leq 1.03)

                           = 1 - 0.8485 = <u>0.1515</u>

Now, at a 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 1.03 < 1.645, <u><em>so we insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

<u>Problem 2:</u>

We are given that a random sample of 35 trading days in 2014 resulted in a  sample mean of 3.28 million shares with a standard deviation of 1.68 million shares.

Let \mu = <em>mean daily volume in Google stock</em>

So, Null Hypothesis, H_0 : \mu = 5.44 million shares    {means that the volume of Google stock has not changed}

Alternate Hypothesis, H_A : \mu \neq 5.44 million shares     {means that the volume of Google stock has changed}

This is a two-tailed test.

The test statistics that would be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                       T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean volume in Google stock = 3.28 million shares

            s = sample standard deviation = 1.68 million shares

           n = sample of trading days = 35

So, <u><em>the test statistics</em></u> =  \frac{3.28-5.44}{\frac{1.68}{\sqrt{35} } }  ~ t_3_4

                                    =  -7.606

The value of t-test statistics is -7.606.

<u>Also, the P-value of the test statistics is given by;</u>

              P-value = P(t_3_4 < -7.606) = Less than 0.05%

Now, at a 0.05 level of significance, the t table gives a critical value of -2.032 and 2.032 at 34 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, <u><em>so we sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the volume of Google stock has changed.

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