A parachutist's rate during a free fall reaches 70 meters per second. What is this rate in feet per second? At this rate, how ma
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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, : p
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, : 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. = ~ N(0,1)
where, = sample proportion of adult Americans who were worried about having enough saved for retirement =
= 0.53
n = sample of adult Americans = 295
So, <u><em>the test statistics</em></u> =
= 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 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 = <em>mean daily volume in Google stock</em>
So, Null Hypothesis, :
= 5.44 million shares {means that the volume of Google stock has not changed}
Alternate Hypothesis, :
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. = ~
where, = 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> = ~
= -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( < -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.