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lyudmila [28]
4 years ago
9

A recent article found that Massachusetts residents spent an average of $860.70 on the lottery in 2010, more than three times th

e U.S. average (http://www.businessweek.com, March 14, 2012). A researcher at a Boston think tank believes that Massachusetts residents spend significantly less than this amount. He surveys 100 Massachusetts residents and asks them about their annual expenditures on the lottery.
a. Specify the competing hypotheses to test the researcher’s claim. H0: μ = 860.70; HA: μ ≠ 860.70 H0: μ ≥ 860.70; HA: μ < 860.70 H0: μ ≤ 860.70; HA: μ > 860.70
b. Specify the critical value(s) of the test at the 10% significance level. (Negative value should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round your answer to 3 decimal places.) Critical value rev: 08_20_2013_QC_33738
c. Compute the value of the appropriate test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.) Test statistic d. At α = 0.10, what is the conclusion? Reject H0; the average Massachusetts residents spent less than $860.70 on the lottery in 2010 Reject H0; the average Massachusetts residents spent no less than $860.70 on the lottery in 2010 Do not reject H0; the average Massachusetts residents spent less than $860.70 on the lottery in 2010 Do not reject H0; the average Massachusetts residents spent no less than $860.70 on the lottery in 2010 Annual Lottery Expenditures (in $)
790 594 899 1105 1090 1197 413 803 1069 633 712 512 481 654 695 426 736 769 877 777 785 776 1119 833 813 747 1244 1023 1325 719 1182 528 958 1030 1234 833 745 985 774 1002 561 681 546 777 844 856 785 1289 502 703 334 1140 594 719 1002 943 1025 969 576 627 989 915 662 802 876 962 878 668 1227 947 864 1016 1022 723 665 1072 610 538 992 978 1291 1139 1111 873 850 941 845 639 495 1016 939 974 893 645 1098 788 682 686 764 759
Mathematics
1 answer:
SCORPION-xisa [38]4 years ago
8 0
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Answer:

a. The positive difference between Nelson's height and the population mean is: \\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

b. The difference found in part (a) is 1.174 standard deviations from the mean (without taking into account if the height is above or below the mean).

c. Nelson's z-score: \\ z = -1.1739 \approx -1.174 (Nelson's height is <em>below</em> the population's mean 1.174 standard deviations units).

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Step-by-step explanation:

The key concept to answer this question is the z-score. A <em>z-score</em> "tells us" the distance from the population's mean of a raw score in <em>standard deviation</em> units. A <em>positive value</em> for a z-score indicates that the raw score is <em>above</em> the population mean, whereas a <em>negative value</em> tells us that the raw score is <em>below</em> the population mean. The formula to obtain this <em>z-score</em> is as follows:

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  • Nelson's height is 68 in. In this case, the raw score is 68 in \\ x = 68 in.
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  • \\ \sigma = 2.3in.

With all this information, we are ready to answer the next questions:

a. What is the positive difference between Nelson​'s height and the​ mean?

The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):

\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

That is, <em>the positive difference is 2.7 in</em>.

b. How many standard deviations is that​ [the difference found in part​ (a)]?

To find how many <em>standard deviations</em> is that, we need to divide that difference by the <em>population standard deviation</em>. That is:

\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174

In words, the difference found in part (a) is 1.174 <em>standard deviations</em> from the mean. Notice that we are not taking into account here if the raw score, <em>x,</em> is <em>below</em> or <em>above</em> the mean.

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Using formula [1], we have

\\ z = \frac{x - \mu}{\sigma}

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\\ z = \frac{-2.7\;in}{2.3\;in}

\\ z = -1.1739 \approx -1.174

This z-score "tells us" that Nelson's height is <em>1.174 standard deviations</em> <em>below</em> the population mean (notice the negative symbol in the above result), i.e., Nelson's height is <em>below</em> the mean for heights in the club presidents of the past century 1.174 standard deviations units.

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Carefully looking at Nelson's height, we notice that it is between those z-scores, because:

\\ -2 < z_{Nelson} < 2

\\ -2 < -1.174 < 2

Then, Nelson's height is <em>usual</em> according to that statement.  

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