Is the null hypothesis Ho rejected at the a = 5% level?

Previously, 3% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than 3% today. She randomly selects 145 pregnant mothers and finds that 3 of them smoked 21 or more cigarettes during pregnancy. Test the researcher's statement at the a = 0.1 level of significance. What are the null and alternative hypotheses? Họ: P = 0.03 versus H4: p < 0.03 (Type integers or decimals. Do not round.) Because npo (1- Po) =|| 10, the normal model V be used to approximate the P-value. (Round to one decimal place as needed.)

A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 79%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 79% of married couples. In a random sample of 250 married couples who completed her program, 205 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor's claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H₁. H₂ : D H₁ : 0 (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) 0 (d) Find the p-value. (Round to three or more decimal places.) 0 (e) Is there enough…

A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 77%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 77% of married couples. In a random sample of 215 married couples who completed her program, 167 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)   (a) State the null hypothesis H0 and the alternative hypothesis H1. H0: H1: (b) Determine the type of test statistic to use.   ▼(Choose one)   (c) Find the value of the test statistic. (Round to three or more decimal places.)   (d) Find the p-value. (Round to three…

An institute conducted a clinical trial of its methods for gender selection. The results showed that 153 of 276 babies born to parents using a specific gender-selection method were boys. Use the sign test and a 0.05 significance level to test the claim that the method has no effect on the likelihood of a boy. Let p denote the population proportion of baby boys. Ignore the possibility of the method decreasing the likelihood of a boy. What are the null and alternative hypotheses? O A. Ho: p=0.5 H₂₁: p20.5 O C. Ho: p=0.5 H₁: p=0.5 O B. Ho:p>0.5 H₁: p=0.5 O D. Ho: p=0.5 H₁: p > 0.5 Find the test statistic. Test statistic = Find the P-value. P-value = (Round to four decimal places as needed.) Determine the proper conclusion. Choose the correct answers below. Since the P-value is (Round to two decimal places as needed.) than the significance level, evidence that the method used is effective in increasing the likelihood of a boy. the null hypothesis. There is

A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 78%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 78% of married couples. In a random sample of 240 married couples who completed her program, 189 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.10 level of significance?Perform a one-tailed test. Then complete the parts below.

A hypothesis test on a regression coefficient fails to reject the null hypothesis at the 5% significance level. This means:

Previously, 10.4% of workers had a travel time to work of more than 60 minutes. An urban economist believes that the percentage has increased since then. She randomly selects 75 workers and finds that 16 of them have a travel time to work that is more than 60 minutes. Test the economist's belief at the a = 0.1 level of significance. What are the null and alternative hypotheses? 0.104 versus H1: p Ho: p P = (Type integers or decimals. Do not round.) > 0.104 %3D Because npo (1- Po) 10, the normal model be used to approximate the P-value. %3D (Round to one decimal place as needed.)

A 2012 survey of 417 American adults indicates that 15% of cell phone owners do their browsing on their phone rather than a computer or other device. According to an online article, a report from a mobile research company indicates that 38% of Chinese mobile web users only access the internet through their cell phones. We wish to conduct a hypothesis test to determine if these data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is less than the Chinese proportion of 38%. (a) The Null Hypothesis, in mathematical symbols, is:   H0: p = ______ (b) The Alternative Hypothesis, in mathematical symbols, is:   HA : p = _______ In both the null and alternative hypotheses, the symbol ( p ) is the proportion of American adult cell phone owners who do their browsing on their phone. (c) The standard error based on the null hypothesis is...    :________  (d) Our data indicates a proportion of Americans who use their phones to browse is…

A university has decided to introduce the use of plus and minus grading as long as there is evidence that 60% of the faculty is in support of this change. Write an appropriate null hypothesis and alternate hypothesis to test if the proportion is greater than 60%.