Download 10.2 review answers

Stat and Data Analysis
10.2 review ANSWERS
1) Consumer Reports tested a SRS of 14 brands of vanilla yogurt and found the following numbers of
calories per serving:
160 200 220 230 120 180 140
130 170 190
80 120 100 170
A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What does this
evidence indicate? Does the sample you have indicate that the diet guide is incorrect?
a. Check the conditions (assume the population is normal)
Conditions:
1) Random
2) Normal or N > 30
3) Pop > 10n
1) stated SRS
2) stated “assume population is normal”
3) There are more than 140 brands of yogurt
b. Find the mean and std. deviation
Mean = 157.86
std dev = 44.75
c. Test the claim with a level of significance of 1%.
𝑥̅ = 157.86
s = 44.75
n = 14
α = 0.01
µ = 120
Ho: µ = 120
Ha: µ ≠ 120
𝑡=
157.86−120
44.75⁄
√14
= 3.165
df = 13
2*P(t > 3.165) = 0.0075
We reject Ho b/c p-value of 0.0075 < alpha = 0.01. We have sufficient evidence that the true
average of calories in a serving of vanilla yogurt is not 120 calories.
d. Create a 95% confidence interval for the average calorie content of vanilla yogurt.
157.86 ± (2.16) (44.75⁄
) = (132.02, 183.70)
√14
We are 95% confident that the true average number of calories in a serving of vanilla yogurt is
between 132.02 grams and 183.70 grams.
2) How much time do school-age children spend helping with housework? A random sample of 26
girls in two-parent families where both parents work full-time found a mean of 14.0 minutes per
weekday with a standard deviation of 8.6 minutes. A histogram of the data showed a unimodal
and approximately symmetric distribution (normal). Construct and interpret a 98% confidence
interval for the mean number of minutes spent doing housework.
Conditions:
1) Random
2) N > 30 or normal data
3) Pop > 10n
parents working full time
1) stated Random
2) stated unimodal & symmetric (normal shaped data)
3) There are more than 260 girls in 2 parent families with both
14 ± (2.485) (8.6⁄
) = (9.8086, 18.191)
√26
df = 25
We are 98% confident that the true average amount of minutes spent on housework per day by
children is between 9.8086 mins and 18.191 mins.
3) A study of the ability of individuals to walk in a straight line report the following data on cadence
(strides per second) for a sample of 20 randomly selected healthy men.
0.95 0.85 0.92 0.95 0.93 0.86 1.00 0.92 0.85 0.81
0.78 0.93 0.93 1.05 0.93 1.06 1.06 0.96 0.81 0.96
a. Construct a histogram. Does the data appear to come from a distribution that is unimodal and
symmetric.
The histogram appears unimodal and
approximately symmetric. (normal
shape)
b. Construct and interpret a 96% confidence interval for the mean cadence.
Conditions:
1) Random
2) N > 30 or normal data
3) Pop > 10n
1) Stated Random
2) see histogram above, normal data
3) There are more than 200 healthy men
0.9255 ± (2.539) (0.0809⁄
) = (0.88562, 0.96538)
√20
df = 19
We are 96% confident that the true average cadence is between 0.88562 and 0.96538 strides per
second.
4) A study conducted by researchers at Pennsylvania State University investigated whether time
perception, a simple indication of a person's ability to concentrate, is impaired during nicotine
withdrawal. After a 24-hr smoking abstinence, 20 smokers were asked to estimate how much time
had passed during a 45-sec period. These smokers were not a random sample but were
considered a representative sample. The mean perceived elapsed time was 59.30 seconds with a
standard deviation of 9.84 seconds. Is there evidence that smoking abstinence had a negative
impact on time perception, causing elapsed time to be overestimated? Use a significance level of
1%. You can assume that the data is normally distributed.
Conditions:
1) Random
2) N > 30 or normal data
3) Pop > 10n
HO:   45
HA:   45
t
59.30  45
 6.499
9.84
20
1) Assume representative
2) stated normal data
3) There are more than 200 smokers
x  59.30 s  9.84
n  20 df  19
P(t > 6.499) = 1.582 X 10-6
We reject the Ho b/c the P-Value of 1.582 X 10-6 is less than alpha = 0.01. We have sufficient
evidence that the true average perceived elapsed time for the smokers was greater than 45 seconds.
5) Suppose that the CEO of a large corporation wants to determine whether the average amount of
time spent on personal use of company technology for her employees is greater than the reported
value of 75 minutes per day. Each person in a random sample of 40 employees was contracted and
asked about daily personal use of company technology. The mean time reported was 74.80 minutes
with a standard deviation of 9.45 minutes. Do these data provide evidence that the mean for this
company is greater than 75 minutes? Use a significance level of 5%.
Conditions
1) Random
2) Normal or N > 30
3) Pop > 10n
HO:   75
HA:   75
t
74.80  75
 0.134
9.45
40
1) stated random sample
2) n = 40 > 30
3) There are more than 400 employees in the large corporation
x  74.80 s  9.45
n  40 df  39
P(t > -0.134) = 0.553
We do not reject the Ho, b/c the P-Value of 0.553 is > alpha = 0.05. We do not have sufficient
evidence that the true average amount of time spent on personal use of company technology was
greater than 75 minutes.
6. A confidence interval for the mean is (37.285, 41.315).
(a) Find the sample mean and the margin of error
Sample mean = 𝑥̅ = (37.285 + 41.315)/2 = 39.3
Margin of error = m = 41.315 – 39.3 = 2.015
(c) If the sample size is 50 and the confidence level is 90%, what is the standard deviation?
𝑚 = (𝑡 ∗ ) (𝑠⁄ )
√𝑛
m = 2.015
t* = 1.677 (from INVT program on calculator, using n = 50 and C = 90)
s=?
n = 50
2.015 = (1.677) (𝑠⁄
)
√50
s = 8.496 units
(d) If I decrease my confidence level to 85%, what happens to the: margin of error, confidence
interval, and critical value?
Margin of error = smaller
Confidence interval = narrower
Critical value = smaller
(e) If I increase my sample size to 120, what happens to the: margin of error, confidence interval,
and critical value?
Margin of error = smaller
Confidence interval = narrower
Critical value = smaller