Question

Please help, I asked this question earlier but no one answered it. Will give thumbs up for answer.

2. In New York City in the late 1970s and early 1980s, parking meter collections averaged S1.75 million per month with a standard deviation of 8302 thousand per month. Assume that the amount generated per month is Normal with the mean and standard deviation from the previous sentence For about two years the city hired Brink's Inc. to do collections instead. The data file gives the collections that Brink's reported to the city during that period (a) (2 points) What is the sample size for this data? (b) (4 points) What is the sample mean? (c) (6 points) The city was concerned that Brink's was stealing money from the parking meter receipts. Assume that this sample of months is like a random sample. Given your answers to the first two parts, what is the probability of a sample at least as far below the mean as the sample from the period of the Brink's contract? (d) (2 points) Sketch the correct picture of an appropriately labeled distribution that corresponds to your computation in part (c) 1330143 1525370 1312257 1494717 1385822 1557598 1469066 1547011 1565671 1124576 1773806 1652490 1744558 1678286 1636597 1738847 1576383 1846508 1746899 1685611 1794467 1695017 1671626 1566107

Answer 1

1330143
1525370
1312257
1494717
1385822
1557598
1469066
1547011
1565671
1124576
1773806
1652490
1744558
1678286
1636597
1738847
1576383
1846508
1746899
1685611
1794467
1695017
1671626
1566107
1588310	mean

a)

sample size =n = 24

b)

sample mean = 1588310

c)

The provided sample mean is X-1588310 and the known population standard deviation is σ 302000, and the sample size is n 24 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 1750000 Ha: μ < 1750000 This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used. Based on the information provided, the significance level is α-0.05, and the critical value for a left- tailed test is1.64. The rejection region for this left-tailed test is R-{z : z <-1.64 The z-statistic is computed as follows: X o1588310- 1750000 Since it is observed that z2.623 < z1.64, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p 0.0044, and since p concluded that the null hypothesis is rejected. 0.0044 0.05, it is It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 1750000, at the 0.05 significance level.

D)

Z-Test Results: t-stats2.623, p-value 0.0044 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 4.0 -3.5-3.0 -2.5 -2.0 15 1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0