It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are randomly selected

Question

It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are randomly selected throughout the country, the selections for each chain being independent. The following hourly rates are recorded:
Chain A 4.25 4.75 3.80 4.50 3.90 5.00 4.00 3.80
Chain B 4.60 4.65 3.85 4.00 4.80 4.00 4.50 3.65
Under the assumption of normality and equal variances, can it be concluded at the 5% significance level that chain A pays more than chain B for the job under consideration?

in progress 0
Thu Thảo 2 weeks 2021-07-21T21:09:15+00:00 1 Answers 1 views 0

Answers ( )

    0
    2021-07-21T21:11:00+00:00

    Answer:

    It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

    Step by Step Solution:

    The given data are;

    Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

    Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

    Using the functions of Microsoft Excel, we get;

    The mean hourly rate for fast-food Chain A, \overline x_1 = 4.25

    The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

    The mean hourly rate for fast-food Chain B, \overline x_2 = 4.25625

    The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

    The significance level, α = 5%

    The null hypothesis, H₀:  \overline x_1 = \overline x_2

    The alternative hypothesis, Hₐ:  \overline x_1\overline x_2

    The pooled variance, S_p^2, is given as follows;

    S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}

    Therefore, we have;

    S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682

    The test statistic is given as follows;

    t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}

    Therefore, we have;

    t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176

    The degrees of freedom, df = n₁ + n₂ – 2 = 8 + 8 – 2 = 14

    At 5% significance level, the critical t = 2.145

    Therefore, given that the absolute value of the test statistic is less than the critical ‘t’, we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )