Question

Question 4. According to a major ATM machine company, the mean ATM withdrawal by a sample of 30 people living in the East providence area was found to be $67 while the standard deviation of such ATM withdrawal by all people of this area was found to be $12.

a) For a confidence level of 90%, estimate the mean ATM withdrawal for people living in the East providence area.

b.) It was also found that for a mean ATM withdrawal of $67 by a sample of 30 people, the sample standard deviation was $8. Using this information, for a confidence level of 90%, estimate the mean ATM withdrawal for the people living in East providence. Compare this result with the result obtained in Part (a).

c) On the Thanksgiving day, it was noted that the mean ATM withdrawal by a sample of 30 people was $115 while the sample standard deviation value was $18. For this day, estimate the mean ATM withdrawal for people in the East providence area. Compare your result with that obtained in Part (a)

d) For data given in Part (a) it was found that ATM withdrawal was normally distributed. Calculate the probability that a given person will withdraw an amount greater than $75 on a normal day.

e) For an Error Margin of 2.5, compute the size of sample i) For the case in Part (a). (2) ii) For the case in Part (b).

Answer 1

a)

x̅ = 67, σ = 12, n = 30

90% Confidence interval :

At α = 0.1 two tailed critical value, z_c = ABS(NORM.S.INV(0.1/2)) = 1.645

Lower Bound = x̅ - z_c*σ/√n = 67 - 1.645 * 12/√30 = 63.3963

Upper Bound = x̅ + z_c*σ/√n = 67 + 1.645 * 12/√30 = 70.6037

63.3963 < µ < 70.6037

b)

x̅ = 67, s = 8, n = 30

90% Confidence interval :

At α = 0.1 and df = n-1 = 29, two tailed critical value, t-crit = T.INV.2T(0.1, 29) = 1.699

Lower Bound = x̅ - t-crit*s/√n = 67 - 1.699 * 8/√30 = 64.5183

Upper Bound = x̅ + t-crit*s/√n = 67 + 1.699 * 8/√30 = 69.4817

64.5183 < µ < 69.4817

c)

x̅ = 115, s = 18, n = 30

90% Confidence interval :

At α = 0.1 and df = n-1 = 29, two tailed critical value, t-crit = T.INV.2T(0.1, 29) = 1.699

Lower Bound = x̅ - t-crit*s/√n = 115 - 1.699 * 18/√30 = 109.4161

Upper Bound = x̅ + t-crit*s/√n = 115 + 1.699 * 18/√30 = 120.5839

109.4161 < µ < 120.5839

d)

Null and Alternative hypothesis:

Ho : µ = 75 ; H1 : µ > 75

Test statistic:

z = (x̅- µ)/(σ/√n) = (67 - 75)/(12/√30) = -3.6515

p-value = 1- NORM.S.DIST(-3.6515, 1) = 0.9999

e) E = 2.5

i) Sample size, n = (z * σ / E)² = (1.645 * 12 / 2.5)² = 62.34 = 62

ii) Sample size, n = (z * s / E)² = (1.645 * 8 / 2.5)² = 27.7 = 28