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5) The Ministry of Human Resources wanted to know whether the mean salary of
workers in fast-food industry was larger than that of restaurant industry. To
investigate, the ministry collected the following information on the amounts
earned last week by a sample of workers in fast-food and restaurant industries.
131
135
146
165
136
142
Fast-food
(RM)
102
129
143
149
120
139
Restaurant 130
(RM)
At 0.10 significance level, can we concluded that the mean salary are greater
for the fast-food workers than the restaurant workers? What is the p-value.
(12 marks)​

Respuesta :

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Use the slope-intercept form to find the slope and y-intercept.
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The test statistic of the difference in the mean of the two samples can be

used to determine if the difference in the mean of the samples is significant.

  • There is not enough statistical evidence to conclude that the mean salary for fast-food workers are greater for the fast-food workers than the restaurant workers.

  • The p-value is approximately 0.075

Reasons:

The given data are presented as follows;

[tex]\begin{tabular}{l|c|c|c|c|c|c|c|}Fast-food (RM)&131&135&146&165&136&142&\\Restaurant (RM)&130&102&129&143&149&120&139\end{array}\right][/tex]

From MS Excel, we have;

Mean salary of fast-food industry workers, [tex]\overline x_1[/tex] = 142.5

Standard deviation, s₁ = 12.24337

The sample size, n₁ = 6

Mean salary for restaurant industry workers, [tex]\overline x_2[/tex] = 130.2857

The standard deviation, s₂ = 15.78728

Sample size, n₂ = 7

The null hypothesis is H₀; [tex]\overline x_1[/tex] = [tex]\overline x_2[/tex]

The alternative hypothesis is Hₐ; [tex]\overline x_1[/tex] ≠ [tex]\overline x_2[/tex]

The t-test statistic for the difference between the means is given as follows;

  • [tex]\displaystyle t = \mathbf{\frac{\overline x_1 - \overline x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }}[/tex]

Therefore;

[tex]\displaystyle t = \mathbf{ \frac{142.5 - 130.2857}{\sqrt{\frac{12.24337^2}{6} +\frac{15.78728^2}{7} } }} \approx 1.569[/tex]

The degrees of freedom can be found as follows;

[tex]\displaystyle d.f. = \mathbf{ \frac{\left[\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right]^2 }{\frac{\left(s_1^2/n_1\right)^2}{n_1 - 1} + \frac{\left(s_2^2/n_2\right)^2}{n_2 - 1} }}[/tex]

Using the above formula, we have;

d.f. ≈ 11

From the t-table, the critical-t at 0.10 significance level ≈ 1.80

Given that the test statistic is less than the critical-t, we fail to reject the

the null hypothesis

Therefore;

  • The statistical evidence is not sufficient enough to suggest that there is a difference in the salary of fast-food and restaurant industry workers salary.

The p-value is found from the t-table as follows;

At 10 degrees of freedom, the test statistic of 1.569 is between the alpha levels of 0.1 and 0.05, which gives;

p-value ≈ 0.05 + (0.1 - 0.05)÷2 = 0.075

  • The p-value is approximately 0.075

Learn more about the t-distribution calculations here:

https://brainly.com/question/13844840

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