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Anon25 [30]
3 years ago
13

A researcher collated data on Americans' leisure time activities. She found the mean number of hours spent watching television e

ach weekday to be 2.7 hours with a standard deviation of 0.2 hours. Jonathan believes that his football team buddies watch less television than the average American. He gathered data from 40 football teammates and found the mean to be 2.3. Which of the following are the correct null and alternate hypotheses?
A. H0: u=2.7; Ha: u < 2.7
B. H0: u ≠ 2.7; Ha: u=2.3
C. H0: u=2.7; Ha: u≠ 2.7
D. H0: u=2.7; Ha: u≥ 2.3
Mathematics
2 answers:
Nataly [62]3 years ago
8 0

Answer:

option a

Step-by-step explanation:

did on edge

Semmy [17]3 years ago
5 0

Answer: A. H_0:\mu=2.7\ ;\ H_a:\mu

Step-by-step explanation:

Let \mu be the population mean.

Given : A researcher collated data on Americans' leisure time activities. She found the mean number of hours spent watching television each weekday to be 2.7 hours with a standard deviation of 0.2 hours.

i.e. \mu=2.7

Jonathan believes that his football team buddies watch less television than the average American.

i.e.\mu

Since, the null hypotheses contains equals to sign.

Then, the correct null and alternate hypotheses will be :-

H_0:\mu=2.7\ ;\ H_a:\mu

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Step-by-step explanation:

Given

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Required

A histogram to show the most likely outcome

From the question, we understand that the distribution is binomial.

This is represented as:

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For x = 0 to 5, where x represents the number of free throws; we have:

P(X = x) = ^nC_x * p^x * (1 - p)^{n-x}

P(X = 0) = ^5C_0 * 0.826^0 * (1 - 0.826)^{5-0}

P(X = 0) = ^5C_0 * 0.826^0 * (0.174)^{5}

P(X = 0) = 1 * 1 * 0.000159 \approx 0.0002

P(X = 1) = ^5C_1 * 0.826^1 * (1 - 0.826)^{5-1}

P(X = 1) = ^5C_1 * 0.826^1 * (0.174)^4

P(X = 1) = 5 * 0.826 * 0.000917 \approx 0.0038

P(X = 2) = ^5C_2 * 0.826^2 * (1 - 0.826)^{5-2}

P(X = 2) = ^5C_2 * 0.826^2 * (0.174)^{3}

P(X = 2) = 10 * 0.682 * 0.005268 \approx 0.0359

P(X = 3) = ^5C_3 * 0.826^3 * (1 - 0.826)^{5-3}

P(X = 3) = ^5C_3 * 0.826^3 * (0.174)^2

P(X = 3) = 10 * 0.5636 * 0.030276 \approx 0.1706

P(X = 4) = ^5C_4 * 0.826^4 * (1 - 0.826)^{5-4}

P(X = 4) = 5 * 0.826^4 * (0.174)^1

P(X = 4) = 5 * 0.4655 * 0.174 \approx 0.4050

P(X = 5) = ^5C_5 * 0.826^5 * (1 - 0.826)^{5-5}\\

P(X = 5) = ^5C_5 * 0.826^5 * (0.174)^0

P(X = 5) = 1 * 0.3845 * 1 \approx 0.3845

From the above computations, we have:

P(X = 0)  \approx 0.0002

P(X = 1) \approx 0.0038

P(X = 2)  \approx 0.0359

P(X = 3)  \approx 0.1706

P(X = 4)  \approx 0.4050

P(X = 5) \approx 0.3845

See attachment for histogram

<em>From the histogram, we can see that the most likely outcome is at: x = 4</em>

<em>Because it has the longest vertical bar (0.4050 or 40.5%)</em>

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