The average internet bill in 2017 was $63. In 2020 a random sample of 30 homes...

The average internet bill in 2017 was $63. In 2020 a random sample of 30 homes found the average bill to be $66.12 with a standard deviation of $6.97. can you say with 90% certainty that the internet bills are increasing?

1. state hypotheses

2. state critical value

3. Run the test

4. summarize

Solutions

Expert Solution

As H0 is rejected at 10% los, we conclude that the claim is true that the internet bills are increasing..

Dear student,
I am waiting for your feedback. I have given my 100% to solve your queries. If you satisfied with my answer then please please like this.
Thank You

orchestra answered 1 year ago

Next > < Previous

Related Solutions

The average monthly electric bill of a random sample of 256 residents of a city is...

The average monthly electric bill of a random sample of 256 residents of a city is $90 with a standard deviation of $24. Construct a 95% confidence interval for the mean monthly electric bills of all residents. (Round to two decimal places) [Answer, Answer] Construct a 99% confidence interval for the mean monthly electric bills of all residents. (Round to two decimal places) [Answer, Answer]

In one community, a random sample of 2727 foreclosed homes sold for an average of $440...

In one community, a random sample of 2727 foreclosed homes sold for an average of $440 comma 396440,396 with a standard deviation of $196 comma 981196,981. a) What assumptions and conditions must be checked before finding a confidence interval for the mean? How would you check them? b) Find a 9999% confidence interval for the mean value per home. c) Interpret this interval and explain what 9999% confidence means. d) Suppose nationally, the average foreclosed home sold for $389 comma...

A simple random sample of 60 items resulted in a sample mean of 63. The population...

A simple random sample of 60 items resulted in a sample mean of 63. The population standard deviation is 12. a. Compute the 95% confidence interval for the population mean (to 1 decimal). ( , ) b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals). ( , )

A simple random sample of 60 items resulted in a sample mean of 63. The population...

A simple random sample of 60 items resulted in a sample mean of 63. The population standard deviation is 15. a. Compute the 95% confidence interval for the population mean (to 1 decimal). b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals).

17. In a random sample of 30 business students, the average time to complete a statistics...

17. In a random sample of 30 business students, the average time to complete a statistics exam was 43.4 minutes. Assume that the population standard deviation to complete the exam is 8.7 minutes. The margin of error for an 80% confidence interval around this sample mean is ________.

The average height of an NBA player is 6.698 feet. A random sample of 30 players’...

The average height of an NBA player is 6.698 feet. A random sample of 30 players’ heights from a major college basketball program found the mean height was 6.75 feet with a standard deviation of 5.5 inches. At α = 0.05, is there sufficient evidence to conclude that the mean height differs from 6.698 feet? State the null and alternate hypothesis State the critical value(s). Enter the appropriate letter. Calculate the test value Make the decision by rejecting or not...

In a random sample of 63 professional actors, it was found that 45 were extroverts. (a)...

In a random sample of 63 professional actors, it was found that 45 were extroverts. (a) Let p represent the proportion of all actors who are extroverts. Find a point estimate for p. (Round your answer to four decimal places.) (b) Find a 95% confidence interval for p. (Round your answers to two decimal places.) lower limit upper limit Give a brief interpretation of the meaning of the confidence interval you have found. We are 5% confident that the true...

A random sample of 30 marathon runners demonstrated that the average age was 38 years old,...

A random sample of 30 marathon runners demonstrated that the average age was 38 years old, with s = 5 years. Find a 95% confidence interval for the average age of marathon runners.

a)In order to determine the average price of hotel rooms in georgia, a sample of 63...

a)In order to determine the average price of hotel rooms in georgia, a sample of 63 hotels were selected. It was determined that the test statistic (z) was $1.74. We would like to test whether or not the average room price is significantly different from $110. Population standard deviation is known to us.Compute the p-value. b)In order to determine the average price of hotel rooms in Atlanta, a sample of 36 hotels were selected. It was determined that the average...

At Finy Community College, 63% of the student body are freshmen. A random sample of 42...

At Finy Community College, 63% of the student body are freshmen. A random sample of 42 student names taken from the Dean’s Honor List over the past several semesters showed that 28 were freshmen. Does this indicate the population proportion of freshmen on the Dean’s list is greater than 63%? Use a 1% level of significance. (a) state the null hypothesis and (b) the alternate hypothesis, (c) State the type of test that you will use to test the hypothesis,...

Subjects