Unit 6: CIs for Proportions
A researcher wants to estimate the proportion of people in a city who prefer electric cars over gasoline cars. In a random sample of 200 people, 130 prefer electric cars.
1) The sample proportion (p_hat) is ____.
2) The standard deviation of the sample proportion is ____.
1) What is 0.65?
2) What is 0.03?
sqrt(0.65 * 0.35 / 200) = 0.03
Hypothesis testing is a technique by which we use our sample to make an inference about a population _______.
A sample of 50 people has a mean of 35 with a standard deviation of 10. If you were to construct a 95% confidence interval, the margin of error would be _____.
What is 2.77?
= 1.96 * 10 / sqrt(50) = 2.77
A) E(Z) = ____.
B) Var(Z) = ____.
A) What is 13?
2*5 + 3*2 - 3 = 13
B) What is 72?
22 * 9 + 32 * 4 = 72
A statistic describes a sample, whereas a ______ describes a population.
1) If you increase the sample size when constructing a confidence interval, the width of the confidence interval ______.
2) If you increase the confidence level when creating a confidence interval, the width of the confidence interval ______.
1) What is decreases?
higher n --> lower MoE --> narrower CI
2) What is increases?
higher confidence --> higher CV --> higher MoE --> wider CI
I am interested in conducting a hypothesis test for the claim that the proportion of all University of Iowa student who like Mickey's is equal to 50%. When sampling 1000 students, the proportion who like Mickey's was 55%. The standard error used in the hypothesis test is _____.
What is 0.05?
standard error = sqrt(0.5 * 0.5 / 100) = 0.05
*remember to use p0 when finding the standard error (we assume the null to be true).
Given that alpha = 0.05 and n = 45, the critical value for a two-sided t-test is ____.
What is t.inv(0.025, 44)?
In a study, 120 out of 200 men and 90 out of 180 women support a new policy. For simplicity, let p_hat1 be the proportion of men who support the policy and let p_hat2 be the proportion of women who support the policy. A 95% confidence interval for the difference in proportions of support for the policy is _____.
What is (0.0003, 0.1997)?
p_hat1 = 0.60 | n1 = 200 | p_hat2 = 0.50 | n2 = 180 | p_hat1 - p_hat2 = 0.10
standard error = sqrt(0.6 * 0.4 / 200 + 0.5 * 0.5 / 180) = 0.0509
CI: 0.10 +/- 1.96 * 0.0509 = (0.0003, 0.1997)
In a quality control test, a factory has a 90% success rate in producing defect-free products. A batch contains 12 products. Let X represent the number of defect-free products in a batch. What is the expected number of defect-free products in the batch?
What is 10.8?
Binomial E(x) = n*p = 12 * 0.9 = 10.8
A researcher is studying the average number of hours college students spend exercising monthly. In a sample of 36 students, the average time spent exercising monthly was 15 hours with a standard deviation of 4 hours. Of the same sample, 10.8% of students indicated their favorite day to exercise was on Sunday.
sx_bar - sp_hat = ____
What is 0.616?
sx_bar = 4 / sqrt(36) = 0.666
sp_hat = sqrt(0.108 * (1 - 0.108) / 36) = 0.05
0.666 - 0.05 = 0.616
If Ha: p < 0.50 and alpha = 0.10, the critical value is _____.
Answer with an Excel formula.
What is norm.inv(0.10, 0, 1)?
A company produces lightbulb. A random sample of 100 of the company's bulb has a mean lifespan of 1500 hours with a standard deviation of 200 hours. A 98% confidence interval for the true population mean lifespan of the bulbs is _____.
t.inv(0.025, 100) = 2.23
t.inv(0.02, 99) = 1.98
t.inv(0.01, 99) = -2.36What is (1452.71, 1547.29)?
1500 +/- 2.36 * (200 / sqrt(100)) = (1452.71, 1547.29).
A researcher compares test scores from two groups:
Group A: n = 30, x_bar = 78, sx_bar = 10
Group B: n = 35, x_bar = 74, sx_bar = 12
A 95% confidence interval for the difference in population means is _____.
t.inv(0.025, 35) = 1.88What is (-1.58, 9.58)?
CV = 2.045
standard error = sqrt(102 / 30 + 122 / 35) = 2.729
CI: (78 - 74) +/- 2.045 * 2.729 = (-1.58, 9.58)
You are interested in predicting the annual sales of a grocery store based on the store's square footage. The LSQ equation that results from your regression is:
y_hat = 235,00 + 124.40x
If the actual sales for a grocery store of 10,000 square feet is $1,350,000, what is the residual and is your model over or underestimating?
What is $102,650 and underestimating?
residual = 1,350,000 - 1,247,350 = 102,650
Since the residual is positive, the actual value is higher than the predicted.
Of a sample of 345 Statistics for Business students, 36% said Probability was the hardest unit to understand. The 90% confidence interval for the true proportion of Statistics for Business students who think Probability is the hardest unit to understand is _____.
What is (0.317, 0.403)?
p_hat = 0.36 | n = 345 | 90% CV: 1.645
CI = 0.36 +/- 1.645 * sqrt(0.36 * 0.64 / 345 = (0.317, 0.403)
On March 28th, 2016, a crossover episode of Flash with Supergirl aired on CBS reaching a different audience. The CW claims that, due to the crossover episode, more than 34.8% of viewers watch Flash in their television lineup. They conducted a random survey of 350 viewers and found that 137 viewers watch Flash. Test CW's claim at alpha = 0.05.
What is fail to reject the null hypothesis?
H0: p = 0.348 | Ha: p > 0.348 | p_hat - 0.3914 | n = 350
standard error = sqrt(0.348 * (1 - 0.348) / 350) = 0.0255
test statistic = (0.3914 - 0.348) / 0.0255 = 1.702
Since 1.702 < 1.96, fail to reject the null.
A chocolate company claims that the mean weight of its chocolate bars is 100 grams. A random sample of 36 chocolate bars has a mean weight of 98 grams with a standard deviation of 5 grams. Test, at alpha = 0.05, whether the true average weight of the company's chocolate bars differs form 100 grams.
t.inv(0.025, 35) = 2.03
t.inv(0.05, 35) = 1.98
t.inv(0.025, 36) = 2.31
CV = 2.03
test statistic = (98 - 100) / (5 / sqrt(100)) = -2.4
Since |-2.4| > 2.03, we reject the null hypothesis.
A nutritionist wants to determine if there is a difference in the average daily calorie intake between two groups. A sample of 15 vegans has a mean intake of 1800 kcal with a standard deviation of 200 kcal. A sample of 20 non-vegans has a mean intake of 2100 kcal with a standard deviation of 250 kcal.
At alpha = 0.05, test whether there is a significant difference in the mean daily calorie intake between these two groups.
x_bar1 = 1800 | n1 = 15 | x_bar2 = 2100 | n2 = 20 | CV = 2.145
standard error = sqrt( 2002 / 15 + 2502 / 20) = 76.103
test statistic = (1800 - 2100) / 76.103 = -3.94
Since | -3.95 | > 2.145, reject the null.
Imagine you are the line manager at a very large factory. Assume each product is either defective or not defective. The non-defective rate for each product is 88%. The probability the first defective productive is found between the 5th and 9th products (inclusive) is _____.
What is 0.2832?
p = 1 - 0.88 = 0.12
P(X<=9) = 1 - (1 - 0.12)|9| = 0.6835
P(X<=4) = 1 - (1 - 0.12)|4| = 0.4003
P(5<=X<=9) = 0.6835 - 0.4003 = 0.2832