Question

In: Statistics and Probability

A simple random sample was taken of 1000 shoppers Respondents were classified by gender (male or...

A simple random sample was taken of 1000 shoppers Respondents were classified by gender (male or female) and by meat department preference (beef, chicken, fish). Results are shown in the contingency table: Gender: Beef Chicken Fish Row Total Male 200 150 50 400 Female 250 300 50 600 Column Total 450 450 100 1000 Is there a difference in meat preference between males and females? The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. Use a 0.05 level of significance. Χ2 = (200 - 180)2/180 + (150 - 180)2/180 + (50 - 40)2/40 + (250 - 270)2/270 + (300 - 270)2/270 + (50 - 60)2/60 Χ2 = 400/180 + 900/180 + 100/40 + 400/270 + 900/270 + 100/60 Χ2 = 2.22 + 5.00 + 2.50 + 1.48 + 3.33 + 1.67 = 16.2 As you can see, the math gets a bit cumbersome, so I did it for you! h. What will the distribution look like on a Bell curve? I need help with letter H. I understand all the rest.

Solutions

Expert Solution

if any quire's please comment below and rata by thumb up Thank you;


Related Solutions

A simple random sample was taken of 1000 shoppers Respondents were classified by gender (male or...
A simple random sample was taken of 1000 shoppers Respondents were classified by gender (male or female) and by meat department preference (beef, chicken, fish). Results are shown in the contingency table:         Gender: Beef Chicken Fish Row Total Male 200 150 50 400 Female 250 300 50 600 Column Total 450 450 100 1000 Is there a difference in meat preference between males and females? The solution to this problem takes four steps: (1) state the hypotheses, (2)...
A public opinion poll surveyed a simple random sample of 1000 voters. Respondents were classified by...
A public opinion poll surveyed a simple random sample of 1000 voters. Respondents were classified by gender (male or female) and by voting preference (Republican, Democrat, or Independent). Results are shown in the contingency table below.                                                 VOTING PREFERENCES                                                 Rep        Dem      Ind         Row Total Male                                      350         075         25           450 Female                                 275         240         35           550 Column Total                     625         315         60           1000 Is there a gender gap? D3 the men's voting preferences differ significantly from the women's preferences? Use...
Measurements of length (cm) were taken for a sample of fish. The gender (male or female)...
Measurements of length (cm) were taken for a sample of fish. The gender (male or female) of each fish was also recorded. The most appropriate technique to explore whether there is a difference in the mean weights of fish between males and females would be: Select one: a. 1 sample z-test b. testing for two proportions c. 2 sample t-test. d. 1 sample t-test
Suppose a simple random sample of athletes in the NBA heights is taken. There were 28...
Suppose a simple random sample of athletes in the NBA heights is taken. There were 28 athletes in the sample with a mean height of 78.4 inches and standard deviation of 2 inches. It has been confirmed through statistical analysis that NBA player heights follows a normal distribution. a. what parameter are we estimating? b. Explain the requirements as they relate to the problem c.What is the point estimate of the parameter? d.Input the margin of error for a 95%...
When birth weights were recorded for a simple random sample of 16 male babies born to...
When birth weights were recorded for a simple random sample of 16 male babies born to mothers taking a special vitamin supplement, the sample had a mean of 3.675 kilograms and a standard deviation of 0.657 kilogram. The birth weights for all babies are assumed to normally distributed. Use a 0.05 significance level to test the claim that the mean birth weight for all male babies of mothers taking the vitamin supplement is different from 3.39 kilograms, which is the...
Pulse rates for Men A simple random sample of 40 pulse rates of women were taken....
Pulse rates for Men A simple random sample of 40 pulse rates of women were taken. The mean of women’s pulse rates have a mean of 67.3 beats per minute and a standard deviation of 10.3 beats per minute. Construct a 99% confidence interval estimate of the standard deviation of the pulse rates of men. Pulse rates are normally distributed.
A poll classified respondents by gender and political​ party, as shown in the table. We wonder...
A poll classified respondents by gender and political​ party, as shown in the table. We wonder if there is evidence of an association between gender and party affiliation.​ (Assume a significance level of alpha=0.05​) D R I Male 36 45 25 Female 51 38 13 calculate chi-square statistic x²=
A public opinion poll surveyed a simple random sample of 550 voters in Oregon. The respondents...
A public opinion poll surveyed a simple random sample of 550 voters in Oregon. The respondents were asked which political party they identified with most and were categorized by residence. Results are shown below. Decide if voting preference is independent from location of residence. Let α=0.05. Republican Democrat Independent NW Oregon 85 103 22 SW Oregon 45 66 10 Central Oregon 46 53 9 Eastern Oregon 67 33 11 Enter the test statistic - round to 4 decimal places.
A simple random sample of 1000 elements generates a sample proportion p =0.68   a. Provide the...
A simple random sample of 1000 elements generates a sample proportion p =0.68   a. Provide the confidence interval for the population proportion (to 4 decimals). , (     , ) b. Provide the confidence interval for the population proportion (to 4 decimals). , ( , )
A simple random sample of 400 persons is taken to estimate the percentage of Republicans in...
A simple random sample of 400 persons is taken to estimate the percentage of Republicans in a large population. It turns out that 210 of the people in the sample are Republicans. True or False and explain. a. The sample percentage is 52.5%; the SE for the sample percentage is 2.5%. b. 52.5% ± 2.5% is a 75%-confidence interval for the population percentage. c. 52.5% ± 5% is a 95%-confidence interval for the sample percentage. d. 52.5% ± 5% is...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT