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## Question 1

Consider that for years it has been accepted that the mean of population 1 is the same as the mean of population 2, but that now population 1 is believed to have a higher mean than population 2. Letting α = .05 and assuming the populations have equal variances and are normally distributed, use the following data to test this belief.

 Sample 1 Sample 2 41.6 43.8 41.1 36.4 46.0 51.0 41.2 41.3 44.4 45.4 44.1 39.8 41.6 45.7 36.5 42.3 48.3 46.0 41.0 41.2

Question 2

Mitra company has been averaging 17.9 orders per week for several years. However, during a recession, weekly orders have appeared to slow. Suppose, John Dong, Mitra’s production manager randomly samples 32 weeks during the recession period and finds a sample mean of 15.5 orders per week. The population standard deviation is 2.25 orders per week. Test to determine whether the average number of weekly orders decreased in the recession period, using a = .10. How would the result would change if a =. 05

Question 3

Suppose Tom wants to determine whether the average values for populations 1 and 2 are different, and you randomly gather the following data.

1. Test Tom’s conjecture, using a probability of .01 of committing a Type I error. Assume the population variances are the same and x is normally distributed in both populations.
1. Use these data to construct a 98% confidence interval for the difference between the two population means.
 Sample 1 Sample 2 9 1 2 8 8 0 12 7 11 10 9 8 2 10 2 5 7 8 8 10 10 9 11 9 11 2 3 9 4 5 10 8 10 11 10 7

Question 4

A group of Malaysian investors wants to develop a chain of fast-food restaurants in Australia. In determining potential costs for each facility, they must consider, among other expenses, the average monthly electricity bill. The investors decide to sample some fast-food restaurants currently operating to estimate the monthly cost of electricity. They want to be 99% confident of their results and want the error of the interval estimate to be no more than \$100. They estimate that such bills range from \$600 to \$2500. How large a sample should they take? How would the result change if the Malaysian investors want to be 95% confident

Question 5

Suppose Samantha has been following a particular banking stock for many years. Samantha is interested in determining the average daily price of this stock over a 10-year period and she has access to the stock reports for these years. However, Samantha does not want to average all the daily prices over 10 years because there are more than 2500 data points, so she decides to take a random sample of the daily prices and estimate the average.

She wants to be 90% confident of his results, Samantha wants the estimate to be within \$2.00 of the true average, and she believes the standard deviation of the price of this stock is about \$12.50 over this period of time. How large a sample should she take? How large a sample should Samantha take if he wants to be 95% confident of her results?

Question 6

Mong & Kong, a reputed bio fuel producer based in Dream Island wishes to determine the proportion of cars that fill up using biofuel.  This population proportion needs to be estimated to within 3% and with 98% level of confidence. Assuming there is no information regarding this population proportion, how large a sample should be selected? If this population proportion needs to be estimated to within 5% and with 95% level of confidence, how large a sample should be selected

## Question 2

Question 7

Across Melbourne, families (of two adults and two children) currently have an average weekly shopping bill at Woolworths of \$253. However, Trisha believes the weekly bill for this type of family in Suburb Ever Green is more than the national average figure. Trisha decides to test his assumption by randomly selecting 24 families (of two adults and two children) who shop at Woolworths in Ever Green area.

The data collected is presented in the table below. Assuming the bill amounts for families shopping at this supermarket chain are normally distributed, and using a 10% level of significance, is there evidence to support his belief? How would the result change using a 5% level of significance

 220 234 225 212 241 255 249 262 288 295 289 270 296 304 299 315 323 331 328 320 334 296 280 292

Question 8

Leighton collected the following data from a random sample of 11 items.

 1175 1200 1080 1400 1225 1280 1090 1387 1275 1287 1201

Use these data and a 5% level of significance to test the following hypotheses, assuming that the data come from a normally distributed population.

H0: m £ 1158

Ha: m > 1158

How would the result change using a 2% level of significance

Question 9

A survey conducted by ANZ Bank claimed that 79% of their customer’s aged over 65 used computers at home to make regular bill payments online for things such as electricity, gas and telephone. All Graduate Research, an independent research company based in Wollongong decided to test this claim. A random sample of 150 ANZ Bank customers over age 65 was selected with 112 found to make regular bill payments online. Using a 5% level of significance, is there sufficient evidence to conclude that the proportion is significantly less than the 79% claimed by the banking industry? How would the result change using a 1% level of significance

Question 10

The chief data analyst of Dynamic Fashion Company believes the per diem cost in Darwin rose significantly between 2009 and 2015. To test this belief, the auditor samples 51 business trips from the company’s records for 2009; the sample average was \$192 per day, with a population standard deviation of \$18.70. The chief data analyst selects a second random sample of 47 business trips from the Dynamic Fashion Company’s records for 2015; the sample average was \$198 per day, with a population standard deviation of \$15.10. If he uses a risk of committing a Type I error of .02, does the auditor find that the per diem average expense in Adelaide has gone up significantly

Question 11

A team of researchers at Flinders Business has found that the average number of lead particles of air in Adelaide City is currently 82 micrograms per cubic metre. Suppose the Adelaide City Council officials work with businesses, commuters and industries and introduce initiatives to try to reduce this figure. The Adelaide City Council hires an environmental company to take random measurements of suspended particles, in micrograms per cubic metre of air, over a period of several weeks. The resulting data follow. Take the population standard deviation to be 9.90 micrograms per cubic metre of air. Use these data to determine whether the city’s average number of suspended particles is significantly lower than it was when the initial measurements were taken. Let a =.05. How would the result change using a =.01

 64 65 64 66 63 68 66 66 71 74 72 74 71 75 75 75 76 77 76 77 76 78 78 81 79 80 79 80 78 82 80 78 88 89 88 89 85

Question 12

Garry Anderson, a consultant for Love My Life Insurance Company conducted a survey of insurance policy holders and found that 46.9% of them read their policy document when it is renewed each year. Love My Life Insurance Company decided to invest considerable time and money into rewriting their policies to make them easier for their policy holders to read. After using the newly written policies for one year, Love My Life Insurance Company managers wanted to determine if .

the rewriting had significantly changed the proportion of policyholders who read their insurance policy upon renewal. They decided to randomly contact 382 of the company’s policy holders who renewed a policy in the past year and ask them whether they had read their current policy; 164 responded that they had read the policy. Use a 5% level of significance to test the appropriate hypothesis. How would the result change if a 1% level of significance is used to test the appropriate hypothesis.

Question 1

Population = Equal variance

Null hypothesis: Mean of population 1 is not higher than mean of population 2.

Alternative hypothesis: Mean of population 1 is higher than mean of population 2.

The two sample t test for equal variance

The one tailed p value comes out to be 0.0003 which is lower than significance level and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Thus, it can be said that mean of population 1 is higher than mean of population 2.

Question 2

Population standard deviation = 2.25 orders per week

Sample size = 32

Null hypothesis: Average number of weekly orders is not decreased in recession period.

Alternative hypothesis: Average number of weekly orders is decreased in recession period.

It is apparent from the above that population standard deviation is given and sample size is higher than 30 and therefore, as per Central Limit Theorem the z statistic would be used to check the validity of the claim.

Average order = 17.9 orders per week

Sample mean = 15.5 orders per week

The p value = 0.00 (standard normal table)

Alpha =0.05

The p comes out to be 0.00 which is lower than significance level and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, the average number of weekly orders is decreased in recession period.

Question 3

• Probability level = 0.01

Null hypothesis: Average of population 1 and population 2 is same.

Alternative hypothesis: Average of population 1 and population 2 is different.

The two sample t test for equal variance

The two tailed p value comes out to be 0.000 which is lower than significance level and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Thus, it can be said that average of population 1 and population 2 is different.

• 98% confidence interval

Question 4

Margin of error would be 100

Further,

Margin of error = z value * Standard deviation / sqrt (n)

The z value for 99% confidence = 2.58

With the help of 6 sigma estimation of standard deviation

Margin of error = z value * Standard deviation / sqrt (n)

Therefore, the sample size would be 67.

Question 5

Margin of error = \$2

Standard deviation = \$12.50

Z value for 95% confidence = 1.96

Margin of error = z value * Standard deviation / sqrt (n)

Minimum sample size = (z value * Standard deviation/ Margin of error)^2

Margin of error = 3%

The z value for 99% confidence = 2.58

(Assuming)

As the population standard deviation is not given and thus, t stat would be used to check the claim.

Degree of freedom = 24-1 = 23

The p value (two tailed) = 0.0007

Significance level = 10% and 5%

The two tailed p value comes out to be 0.000 which is lower than significance level 5% as well as 10% and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, it can be said that average weekly bill of Suburb Ever Green is more than the national average bill (\$253).

Significance level = 5%

Number of observation = 24

As the population standard deviation is not given and thus, t stat would be used to check the claim.

Degree of freedom = 11-1=10

The p value (one tailed) = 0.0156

Significance level = 5%

The one tailed p value comes out to be 0.0156 which is lower than significance level 5% and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, it can be seen that mean is higher than 1158.

Significance level = 2%

The one tailed p value comes out to be 0.0156 which is lower than significance level 2% and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, it can be said that mean is higher than 1158.

Sample size n= 150

The p value (one tail) = 0.098

Significance level = 5% and 1%

The one tailed p value comes out to be 0.098 which is higher than significance level 5% and  1% and hence, insufficient evidence is present to reject the null hypothesis. Therefore, it can be said that proportion is not significantly lower than 79% as claimed by banking industry.

Significance level = 0.02

The one tailed p value comes out to be 0.0427 which is higher than significance level 0.02 and hence, insufficient evidence is present to reject the null hypothesis. Therefore, it can be said that per diem average expense in Adelaide has not gone up significantly.

Population standard deviation = 9.9 microgram per m^3 of air

Sample size = 37

It is apparent from the above that population standard deviation is given and sample size is higher than 30 and therefore, as per Central Limit Theorem the z statistic would be used to check the validity of the claim.

Sample mean = 75.757 microgram per m^3 of air

Standard deviation = 7.205 microgram per m^3 of air

The p value = 0.00 (standard normal table)

Alpha =0.05 and 0.01

The p comes out to be 0.00 which is lower than significance level and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, it can be said that average number of lead particle of air in Adelaide City is   higher than 82 microgram per m^3 of air.

Sample size n= 382

The p value (two tail) = 0.06

Significance level = 5% and 1%

The two tailed p value comes out to be 0.098 which is higher than significance level 5% and 1% and hence, insufficient evidence is present to reject the null hypothesis. Therefore, it can be said that proportion is not equal to 46.9%.

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