Question

In: Math

The data below represents the amount that a sample of fifteen customers spent for lunch ($)...

  1. The data below represents the amount that a sample of fifteen customers spent for lunch ($) at a fast-food restaurant:

8.42   6.29   6.83   6.50   8.34   9.51   7.10   6.80   5.90   4.89   6.50   5.52   7.90   8.30   9.60

  • At the 0.01 level of significance, is there evidence that the mean amount spent for lunch is different from $6.50? Follow and show the 7 steps for hypothesis testing.  
  • Determine the p-value and interpret its meaning.
  • What assumption must you make about the population distribution in order to conduct the test in part a? Is the assumption valid? Use and include an appropriate graph from Minitab. Write a couple of sentences supporting your answer.  
  • Verify your results using Minitab by attaching or including the Minitab output.

2. Of 900 surveyed active LinkedIn members, 368 reported that they are planning to spend at least $1,000 on consumer electronics in the coming year.

  • At the 0.05 level of significance, is there evidence that the percentage of all LinkedIn members who plan to spend at least $1,000 on consumer electronics is greater than 38%? Follow and show the 7 steps for hypothesis testing.
  • Determine the p-value and interpret its meaning.
  • What assumption must you make in order to conduct the test in part a? Demonstrate if the assumption is met.  
  • Verify your results for parts a and b using Minitab. Attach or include your output.  

Solutions

Expert Solution

1. For the testing problem, population distribution is assumed to be normal.

Testing the validity of the assumption, using Minitab (Stat -> Basic Statistics -> Normality Test), we get the following output -

Since P-value = 0.584 > 0.05, so at 5% level of significance, we can conclude that population distribution is normal.

Let denotes the mean amount spent for lunch.

Verify the results using Minitab,


Related Solutions

4. Below is the amount that a sample of 15 customers spent for lunch ($) at...
4. Below is the amount that a sample of 15 customers spent for lunch ($) at a fast-food restaurant: 7.42 6.29 5.83 6.50 8.34 9.51 7.10 5.90 4.89 6.50 5.52 7.90 8.30 9.60 6.80 Recall the lunch at a fast-food restaurant problem from Assignment 4. Let µ represent the population mean amount spent for lunch ($) at a fast-food restaurant. Previously you calculated the mean and standard deviation of the fifteen sample measurements to be x ̅ = $7.09 and...
The table below contains the amount that a sample of nine customers spent for lunch ($)...
The table below contains the amount that a sample of nine customers spent for lunch ($) at a fast-food restaurant: 4.88 5.01 5.79 6.35 7.39 7.68 8.23 8.71 9.88 a.Compute the sample mean and sample standard deviation of the amount spent for lunch. b.Construct a 99% confidence interval estimate for the population mean amount spent for lunch ($) at the fast-food restaurant, assuming the population is normally distributed. c.Interpret the interval constructed in (b). Better to use the computer text...
The table below contains the amount that a sample of nine customers spent for lunch ($)...
The table below contains the amount that a sample of nine customers spent for lunch ($) at a fast-food restaurant: 4.88 5.01 5.79 6.35 7.39 7.68 8.23 8.71 9.88 a. Compute the sample mean and sample standard deviation of the amount spent for lunch. b. Construct a 99% confidence interval estimate for the population mean amount spent for lunch ($) at the fast-food restaurant, assuming the population is normally distributed. c. Interpret the interval constructed in (b).
2. The table below contains the amount that a sample of nine customers spent for lunch​...
2. The table below contains the amount that a sample of nine customers spent for lunch​ ($) at a​ fast-food restaurant. Complete parts a and b below. 4.26 5.23 5.67 6.28 7.44 7.51 8.25 8.69 9.61 Construct a 99% confidence interval estimate for the population mean amount spent for lunch at a fast-food restaurant, assuming normal distribution. The % confidence interval estimate is from $____________to $___________ 3. he table below contains a certain social media​ company's penetration values​ (the percentage...
The data table below contains the amounts that a sample of nine customers spent for lunch​...
The data table below contains the amounts that a sample of nine customers spent for lunch​ (in dollars) at a​ fast-food restaurant 4.23 5.08 5.96 6.45 7.43 7.54 8.36 8.51 9.76 a) At the 0.05level of​ significance, is there evidence that the mean amount spent for lunch is different from ​$6.50​? b) Calculate test stastics ( four decimals places) C) Calculate P value and explain interpretations ( four decimal places)? d) Because the sample size is​ 9, do you need...
The file FastFood contains the amount that a sample of fif- teen customers spent for lunch...
The file FastFood contains the amount that a sample of fif- teen customers spent for lunch ($) at a fast-food restaurant: 7.42 6.29 5.83 6.50 8.34 9.51 7.10 6.80 5.90 4.89 6.50 5.52 7.90 8.30 9.60 At the 0.05 level of significance, is there evidence that the mean amount spent for lunch is different from $6.50? Determine the p-value in (a) and interpret its meaning. What assumption must you make about the population distribu- tion in order to conduct the...
Data were collected on the amount spent by 64 customers for lunch at a major Houston...
Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. Based upon past studies the population standard deviation is known with $6. Round your answers to 2 decimal places. Use the critical value with 3 decimal places. At 99% confidence, what is the margin of error? Develop a 99% confidence interval estimate of the mean amount spent for lunch.   Amount 20.50 14.63 23.77 29.96 29.49 32.70 9.20 20.89 28.87 15.78 18.16 12.16 11.22...
Data were collected on the amount spent for lunch by 64 customers at a major Houston...
Data were collected on the amount spent for lunch by 64 customers at a major Houston restaurant. The sample provided a sample mean of $21.5. Based upon past studies the population standard deviation is known with σ = $6. Develop a 99% confidence interval estimate of the mean amount spent for lunch.
The data below represents the amount of grams of carbohydrates in a sample serving of breakfast...
The data below represents the amount of grams of carbohydrates in a sample serving of breakfast cereal. 10 18 24 30 19 22 24 20 18 25 20 22 19 what is the variance?
A random sample of 29 lunch customers was taken at a restaurant The average amount of...
A random sample of 29 lunch customers was taken at a restaurant The average amount of time the customers in the sample stayed in the restaurant was 45 minutes with a standard deviation of 14 minutes. a) Compute the standard error of the mean? b) Construct a 68% confidence interval for the true average amount of time customers spent in the restaurant? c) Construct a 90% confidence interval for the true average amount of time customers spent in the restaurant?...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT