Question 2.

The 90% confidence interval would be 52.096 to 80.012.

Analyzing the boostrap output for mileage of a random sample of 25 mustang cars we can say In the above random sample of 25 mustag cars, it is plausible to conclude with 90% confidence interval (between 5th percentile and 95 percentile) that the value of the population mean, u will lie in range of 52.096 an 80.012. Since 52 falls out of this range, it is thus the outlier and hence cannot be a plausible population mean (u).

Question 3.

The p-value in support of the alternate hypothesis

a) 0.15

b) 0.85

c) 0.009

d) 0.05

The greater the p-value, the higher the likelihood of rejecting the null hypothesis. Thus, if at all the Ho is to be accepted as true, the p-value of a statistical hypothesis test is the probability of getting a value of the test statistic as extreme as or more extreme than that observed. In this case, the p-value of 0.009 renders more support of the alternate hypothesis.

Question 4.

In a sample of 56 addicts, 28 were given a new drug to help them overcome their addiction, while the remaining 28 were given the current drug. The following is the output from a randomization test for the difference in the proportion of addicts suffering relapses.

What is the difference in proportions in the original data?

The difference in proportions is 0.42857 since the proportion of addicts which took the new drug had 0.3214 relapses, while the proportion that took the old drugs was 0.5354. This difference is shown on the upper right. The -0.2143 on the lower right is the difference in the 3000th resample.

b) What is the p-value and what does it mean in terms of this specific problem?

The probability value (p-value) of a statistical hypothesis test is the probability of getting a value of the test statistic as extreme as or extreme than that observed, when the null hypothesis Ho is true. The p-value is 0.37% and it helps ascertain whether the differences between effects of drugs are in coincidence with Ho or if the proportion of relapses are significantly different due to the new treatment Ha. The p-value of 0.0037 supports the alternate hypothesis and this implies that there are significant differences in the treatments. The probability of getting samples higher than +.4286 or less than -.4286is only 0.37%.

5. Construct a bootstrap distribution of the credit card debt data from your sample using 3000 resamples.

a)

Statistic: mean (CC Debt)

Observed N Mean Std. dev. 2.5th per. 5th per. 50th per. 95th per. 97.5th per.

3501.9 3000 3509.4314 542.57363 2500.8 2645.4833 3485.65 4438.1833 4625.7667

b) What is the mean of your original sample?

Summary statistics:

Column n Mean Variance Std. dev. Std. err. Median Range Min Max Q1 Q3

CC Debt 30 3501.9 9090355.5 3015.0216 550.46512 2412 11215 0 11215 1345 5124

The mean for the original sample is: 3501.9

And mean for the 3000 resamples is 3509.4314

c) What is the 95% confidence interval estimate of the population mean obtained from your bootstrap distribution?

2.5th percentile = 2500.8

97.5th percentile = 4625.7667

Therefore the 95% confidence interval estimate of the population is:

95% CI = 2500.8 - 4625.7667

Question 6.

Null Hypothesis: HO= There is no difference in the mean amount of credit card debt for males and females

Alternate Hypothesis: HA = There exist a difference in the mean amount of credit card debt for males and females.

Paste the output from your randomization test here.

b) What is the mean difference in credit card debt of the two groups in the original data?

Summary statistics for Randomized difference of means:Group by: Gender

Gender n Mean Variance Std. dev. Std. err. Median Range Min Max Q1 Q3

Female 19 -80.171997 1502018.6 1225.5687 281.16474 -40.033493 5076.0287 -2115.9187 2960.11 -857.78469 804.27273

Male 11 79.914311 1815417 1347.3741 406.24858 358.72249 4291.7225 -1790.512 2501.2105 -1171.8517 762.93301

The mean in the original data is 766.2

Mean of data for male: 79.914311

Mean of data for female: 80.171997

Di +/- of means 766.2

c) P-value

The pvalue is .417. If the null hypothesis that there is no difference in the mean amount of credit card debt for males and females is true, the approximately 41.7% of the time, we would see a difference in mean amount of credit extreme as what we saw in our original example just by chance.

One sample proportion hypothesis test:Outcomes in : Randomized difference of means Success : 0.5Group by: Genderp : Proportion of SuccessesH0 : p = 0.5HA : p 0.5Hypothesis test results:

Gender Count Total Sample Prop. Std. Err. Z-Stat P-value

Female 0 19 0 0.11470787 -4.3588989 <0.0001

Male 0 11 0 0.15075567 -3.3166248 0.0009

d) Explanation

There is not that much evidence to reject the null hypothesis of no difference and since the pvalue is >.05, then you would not reject the null.

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