In 2000, 58% of females aged 15 years of age and older lived alone, according to the U.S. Census Bureau. A sociologist tests whether this percentage is different today by conducting a random sample of 500 females aged 15 years of age and older and finds that 285 are living alone. Compute a hypothesis test at 10% level of significance to test the claim that the percentage is different today.

1) Verify that np0(1 - p0) >10. Where p0 is the assumed population proportion. As stated in the problem, n=500 and p0=0.58 here. Plug those values in and show me the final value to confirm that the inequality holds.

2) State the p-value and explain why you should or should not reject the null hypothesis. Explain, in plain English, what we are rejecting or what we are accepting. Don't just say "reject the null hypothesis" or "fail to reject." State the conclusion in terms of the proportion of women living alone. What proportion are we accepting or what proportion are you not accepting. Testing a Population Mean Read the section 10.3 lecture notes. Again, you'll need to actually read the lecture notes and go through the examples carefully before attempting this written assignment. The lecture notes are not perfect nor meant to be a one-stop shop, but they do need to be a starting point. Do students who learned English as well as another language simultaneously score worse on the SAT Critical Reading exam than the general population of test takers? The mean score among all test takers on the SAT Critical Reading exam is 501. A random sample of 100 test takers who learned English as well as another language simultaneously had a mean SAT Critical Reading score of 485 with a standard deviation of 116. Do these results suggest that students who learn English as well as another language simultaneously score worse on the SAT Critical Reading exam? Use a 0.05 significance level. State the appropriate null and alternative hypotheses.

3) State the p=value and use it to explain whether at the =0.05 level of significance we would reject the null hypothesis or not. Explain, in context, what we are rejecting or what we are accepting. Don't just say "reject" or "fail to reject." State the conclusion in terms of the SAT scores for students who learned English and another language simultaneously.

In a certain country, it is thought that the mean age for a first visit to a dentist is 16 years old. A dentist asks new patients if it is the first time visiting a dentist and records the following ten ages for those that made their first visit to a dentist: 21, 18, 11, 18, 23, 20, 15, 22, 18, 16 Suppose that the population is normally distributed, and a t-test can be performed. Based on this data, test the hypothesis that the mean age for a first visit to a dentist is greater than 16 years old. Use a 0.01 significance level.

4) State the p=value and use it to explain whether at the =0.01 level of significance we would reject the null hypothesis or not. Explain, in plain English, what we are rejecting or what we are accepting. Don't just say "reject" or "fail to reject." State the conclusion in terms of the mean age for a first dentist visit.

A researcher wanted to determine whether bicycle deaths were uniformly distributed over the days of the week. In other words, if you were to randomly select an instance of bicycle fatality, would that fatality have an equal chance of having occurred on any given day of the week? She randomly selected 200 deaths that involved a bicycle, recorded the day of the week on which the death occurred, and obtained the following results (the data are based on information obtained from the Insurance Institute for Highway Safety).

Days of the week	Frequency
Sunday	16
Monday	35
Tuesday	16
Wednesday	28
Thursday	34
Friday	41
Saturday	30

Is there reason to believe that bicycle fatalities occur with equal frequency with respect to day of the week at the =0.05 level of significance? State the appropriate null and alternative hypothesis.

The null hypothesis would be that the probability for each day of the week would be equal.

There are 7 days in the week which would make the probability for any given day 1/7. The alternative hypothesis would be that one day (or more) is different. H0: p0=p1=p2=p3=p4=p5=p6=1/7 H1: At least one of the proportions is different from the others.We need to come up with the expected counts. In other words, we apply the probability of each outcome to the sample size. 1/7 times 200 for each outcome as we are assuming a probability of 1/7 for each outcome and there are 200 trials. (1/7)20028.6 Day of the WeekFrequencyExpected CountSunday1628.6Monday3528.6Tuesday1628.6Wednesday2828.6Thursday3428.6Friday4128.6Saturday3028.6As shown in the table, all expected counts are greater than 5. This is more than enough to confirm that the requirements for the goodness-of-fit test are satisfied. We need to enter these values into our lists. To avoid rounding, it's best to enter 200/7 in each L2 entry here. The calculator will then display the rounded answer:

5) State the p-value and explain how you use it to input conclusion on whether at the =0.05 level of significance we would reject the null hypothesis or not. Explain, in plain English, what we are rejecting or what we are accepting.