Can someone explain how I would solve this? I keep getting it wrong, please help The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. Themales took an average of four English courses with a standard deviation of 0.9. The females took an average of five English courses with a standard deviation of 1.1. Are the means statistically the same? (Use a = 0.05)NOTE: If you are usingStudent's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)O Part (a)O Part (b)O Part (c)O Part (d)O Part (e)O Part (f)What is the p-value? (Round your answerfour decimal places.)Explain what the p-value means for this problem.O If H, is false, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at least 1.O If H, is true, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at least 1.O If Ho is false, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at most 1.O If Ho is true, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at most 1.O Part (g)Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.1/2(p-value)1/2(p-valuep-valueXM - XFXM - Xp1/2(p-value)1/2(p-valuep-valueXM - XpXM - Xp

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Description: Can someone explain how I would solve this? I keep getting it wrong, please help The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and d

The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of four English courses with a standard deviation of 0.9. The females took an average of five English courses with a standard deviation of 1.1. Are the means statistically the same? (Use a = 0.05) NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) O Part (a) O Part (b) O Part (c) O Part (d) O Part (e) O Part () What is the p-value? (Round your answer to four decimal places.) Explain what the p-value means for this problem. O If Ho is false, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at least 1. O If H, is true, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at least 1. O If Ho is false, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at most 1. O If Ho is true, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at most 1. O Part (g) Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value. 1/2(p-value) 1/2(p-value p-value XM- Xp XM - Xp 1/2(p-value) 1/2(p-value p-value XM - Xp XM - Xp

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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Can someone explain how I would solve this? I keep getting it wrong, please help

The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The
males took an average of four English courses with a standard deviation of 0.9. The females took an average of five English courses with a standard deviation of 1.1. Are the means statistically the same? (Use a = 0.05)
NOTE: If you are using
Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
O Part (a)
O Part (b)
O Part (c)
O Part (d)
O Part (e)
O Part (f)
What is the p-value? (Round your answer
four decimal places.)
Explain what the p-value means for this problem.
O If H, is false, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at least 1.
O If H, is true, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at least 1.
O If Ho is false, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at most 1.
O If Ho is true, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at most 1.
O Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.
1/2(p-value)
1/2(p-value
p-value
XM - XF
XM - Xp
1/2(p-value)
1/2(p-value
p-value
XM - Xp
XM - Xp

Expert Solution

Step 1

Given:

The mean number of English courses taken in a two-year time period by male and female college students is believed to be same; that is; $μ_{1} = μ_{2}$

where,

$μ_{1}$ be the mean number of English courses taken in a two-year time period by male.

$μ_{2}$ be the mean number of English courses taken in a two-year time period by female.

$n_{1}$ be the number of males; that is; $n_{1}$ = 29.

$n_{2}$ be the number of females; that is; $n_{2}$ = 16.

${\bar{x}}_{1}$ be the sample mean of English courses taken in a two-year time period by male; that is; ${\bar{x}}_{1}$ =4.

${\bar{x}}_{2}$ be the sample mean of English courses taken in a two-year time period by female; that is; ${\bar{x}}_{2}$ =5.

$s_{1}$ be the sample standard deviation of male; that is; $s_{1}$ =0.9.

$s_{2}$ be the sample standard deviation of females; that is; $s_{2}$ =1.1.

Step 2

The hypothesis that can be formed are:

Null hypothesis (H0): $μ_{1} = μ_{2}$

Alternate Hypothesis (H1): $μ_{1} \neq μ_{2}$

The test here is two tailed t test.

The value of test statistics can be calculated as:

$t = \frac{{\bar{x}}_{1} - {\bar{x}}_{2}}{s_{p} \times \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} = \frac{4 - 5}{0.9744 \times \sqrt{\frac{1}{29} + \frac{1}{16}}} = - 3.295$

where, the value of $s_{p}$ is calculated as:

$s_{p} = \sqrt{\frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2}} = \sqrt{\frac{(29 - 1) 0 . 9^{2} + (16 - 1) 1 . 1^{2}}{29 + 16 - 2}} = \sqrt{0.9495} = 0.9744$

Hence, the value of test statistics is obtained as -3.295.

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