There is evidence that men actually do spend significantly more than women on

# There is evidence that men actually do spend

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There is evidence that men actually do spend significantly more than women on Valentine’s Day. Work Show all your work for the questions below. Question 1 μ = 75, σ = 10, x = 85 z = (x – μ)/σ = (85 – 75)/10 = 1 Question 2 μ = 73, σ = 11 (a) x = 95 z = (x – μ)/σ = (95 – 73)/11 = 2 P(x < 95) = P(z < 2) = 0.9772 (b) x1 = 62 and x2 = 84 z1 = (62 – 73)/11 = -1 and z2 = (84 – 73)/11 = 1 P(62 < x < 84) = P(-1 < z < 1) = 0.6827 Question 3 π = 0.19,n = 950 (a) p = 0.20 z = (p – π)/{√π * (1 – π)/n} = (0.20 – 0.19)/{√0.19* (1 – 0.19)/950} = 0.7857 P(p > 0.20) = P(z > 0.7857) = 0.2160 (b) p1 = 0.18 and p2 = 0.21 z1 = (0.18 – 0.19)/{√0.19* (1 – 0.19)/950} = -0.7857 and z2 = (0.21 – 0.19)/{√0.19 * (1 – 0.19)/950} = 1.5713 P(0.18 < π < 0.21) = P(-0.7857 < z < 1.5713) = 0.7259 Question 4 3.27 3.29 3.16 3.20 3.37 3.20 3.23 3.19 3.20 3.24 3.16 3.07 3.27 3.09 3.35 3.15 3.23 3.14 3.05 3.35 3.21 3.14 3.14 3.07 3.10 For the given data, x-bar = 3.173 and s = 0.069 n = 25 μ = 3.16 s = 0.069 x-bar = 3.173 Hypotheses: Ho: μ ≤ 3.16 Ha: μ > 3.16 Decision Rule: α = 0.01 Degrees of freedom = 25 - 1 = 24 Critical t- score = 2.492159469 Reject Ho if t > 2.492159469 Test Statistic: SE = s/n = 0.069/√25 = 0.0138 t = (x-bar - μ)/SE = (3.173 - 3.16)/0.0138 = 0.942028986 p- value = 0.177782159 Decision (in terms of the hypotheses): Since 0.942028986 < 2.492159469 we fail to reject Ho Conclusion (in terms of the problem): There is no sufficient evidence to reject the claim. The mean price may not be greater than \$3.16. Question 5 n = 67 p = 0.47 p' = 40/67 = 0.597 Hypotheses: Ho: p = 0.47 Ha: p ≠ 0.47 Decision Rule: α = 0.05 Lower Critical z- score = -1.9600 Upper Critical z- score = 1.9600 Reject Ho if |z| > 1.9600 Test Statistic: SE = p (1 - p)/n = √(0.47 * (1 - 0.47)/67) = 0.0610 z = (p'- p)/SE = (0.597014925373134 - 0.47)/0.0609746705424572 = 2.0831 p- value = 0.0372 Decision (in terms of the hypotheses): Since 2.0831 > 1.9600 we reject Ho and accept Ha Conclusion (in terms of the problem): There is t evidence to reject the findings of Robert Half International. It appears that the proportion of CFOs who get their money news from the newspapers is not 47%. Question 6 107.48 125.98 143.61 45.53 90.19 56.35 125.53 80.62 70.79 46.37 83.00 44.34 129.63 75.21 154.22 68.48 93.80 85.84 126.11 For the 9 men, x-bar = 110.917 and s = 28.792 For the 10 women, x-bar = 75.463 and s = 30.530 n1 = 9 n2 = 10 x1-bar = 110.916667 x2-bar = 75.463 s1 = 28.7915925 s2 = 30.5296526 Hypotheses: Ha: μ1 ≤ μ2 Ha: μ1 > μ2 Decision Rule: α = 0.05 Degrees of freedom = 9 + 10 - 2 = 17 Critical t- score = 1.73960672 Reject Ho if t > 1.73960672 Test Statistic: Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = √ (((9 - 1) * 28.7915925228182^2 + (10 - 1) * 30.5296526347746^2)/(9 + 10 - 2)) = 29.72440432 SE = s * {(1 /n1) + (1 /n2)} = 29.7244043229287 * √ ((1/9) + (1/10)) = 13.65742128 t = (x1-bar -x2-bar)/SE = (110.916666666667 - 75.463)/13.6574212771363 = 2.595926855 p- value = 0.00941903 #### You've reached the end of your free preview.

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• Fall '19
• Null hypothesis, Statistical hypothesis testing, reject null
• • • 