## How would you calculate the Chi-Square Distribution for the following survey?

Question How would you calculate the Chi-Square Distribution for the following survey? A recent survey asked “Has the recent outbreak of coronavirus (covid-19) directly impacted your business?” to a sample size of 1,172 which represents the 300,000 small business owners in their database. These employers had between 1-465 employees each. The results were as follow for 3 different dates:March 10th74% said no impact23% said yes, negative impact3% said yes, positive impactMarch 20th19% said no impact76% said yes, negative impact5% said yes, positive impactMarch 30th5% said no impact92% said yes, negative impact3% said yes, positive impact

## The display provided from technology available below results from

Question The display provided from technology available below results from using data for a smartphone carrier’s data speeds at airports to test the claim that they are from a population having a mean less than 4.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Identify the test statistic. _____ (Round to two decimal places as needed.) Identify the P-value. _____ (Round to three decimal places as needed.)

## The breaking strengths of cables produced by a certain manufacturer have a mean, μ

Question The breaking strengths of cables produced by a certain manufacturer have a mean, μ , of 1825 pounds, and a standard deviation of 60 pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, 44 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1850 pounds. Assume that the population is normally distributed. Can we support, at the 0.01 level of significance, the claim that the mean breaking strength has increased? (Assume that the standard deviation has not changed.)Perform a one-tailed test. Then fill in the table below.Carry your intermediate computations to at least three decimal places, and round your responses as specified in the table.The null hypothesis:H0:The alternative hypothesis:H1:The type of test statistic:(Choose one)ZtChi squareFThe value of the test statistic:(Round to at least three decimal places.)The critical value at the 0.01 level of significance:(Round to at least three decimal places.)Can we support the claim that the mean breaking strength has increased?YesorNo

## Obesity in children is a major concern because it puts them at risk

Question Obesity in children is a major concern because it puts them at risk for several serious medical problems. Some researchers believe that a major issue related to this is that children these days spend too much time watching television and not enough time being active. Based on a sample of boys roughly the same age and height, data was collected regarding hours of television watched per day and weight. TV watching (hr)Weight (lb)1.5795.01053.5962.5834.0991.0780.568 Use the regression (the perdition equation) to predict the weight of a child who watches 3.0 hours of TV a day.

## A recent study at a local college claimed that the proportion, p

Question A recent study at a local college claimed that the proportion, p , of students who commute more than fifteen miles to school is no more than 25%. If a random sample of 270 students at this college is selected, and it is found that 82 commute more than fifteen miles to school, can we reject the college’s claim at the 0.05 level of significance?Perform a one-tailed test. Then fill in the table below.Carry your intermediate computations to at least three decimal places and round your answers as specified in the table.The null hypothesis:H0:The alternative hypothesis:H1:The type of test statistic:(Choose one)ZtChi squareFThe value of the test statistic:(Round to at least three decimal places.)The p-value:(Round to at least three decimal places.)Can we reject the claim that the proportion of students who commute more than fifteen miles to school is no more than 25%?YesorNo