PROBABILITY STATISTICS Chapter 6 Test Review 0 0400

PROBABILITY & STATISTICS Chapter 6 Test Review

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The critical value for a 90% confidence level is 1. 645.

A random sample of 125 students has a grade point average with a mean of 2. 94 and with a standard deviation of 0. 77. Construct the confidence interval for the population mean, μ, if the confidence level is 95%. The sample size is ≥ 30, so the Central Limit Theorem tells us the sampling distribution is approximately. We can use the normal distribution. The critical value for a 95% confidence level is 1. 96. The population standard deviation is unknown, however we can use the sample standard deviation in its place.

A random sample of 125 students has a grade point average with a mean of 2. 94 and with a standard deviation of 0. 77. Construct the confidence interval for the population mean, μ, if the confidence level is 95%.

Interpret the confidence interval. We are 95% confident that the interval from 2. 81 to 3. 07 captures the actual population mean.

The standard IQ test has a mean of 100 and a standard deviation of 13. We want to be 97% confident that we are within 3 IQ points of the true mean. Determine the required sample size. First, we need the critical value for the 97% confidence level.

The standard IQ test has a mean of 100 and a standard deviation of 13. We want to be 97% confident that we are within 3 IQ points of the true mean. Determine the required sample size.

The standard IQ test has a mean of 100 and a standard deviation of 13. We want to be 97% confident that we are within 3 IQ points of the true mean. Determine the required sample size. Square both sides Round up to the next whole number. The required sample size is at least 89

Here we use the t-distribution table. First, we need the degrees of freedom. Go to the t-distribution table. Line up the degrees of freedom, 19, with the confidence level, . 90.

Find the value of the margin of error for a 95% confidence level for a sample size of 10 with a standard deviation of 3. 9. Assume the population is approximately distributed. The sample size is less than 30. The population standard deviation is unknown. The population is approximately distributed. We can use the t-distribution. First, we need the degrees of freedom.

Find the value of the margin of error for a 95% confidence level for a sample size of 10 with a standard deviation of 3. 9. Assume the population is approximately distributed. Go to the t-distribution table. Line up the degrees of freedom, 9, with the confidence level, . 95. Use the number from the table.

Find the value of the margin of error for a 95% confidence level for a sample size of 10 with a standard deviation of 3. 9. Assume the population is approximately distributed.

In a random sample of 28 families, the average weekly food expense was $95. 60 with a standard deviation of $22. 50. Determine whether a normal distribution or a tdistribution should be used or whether neither of these can be used to construct a confidence interval. Assume the distribution of weekly food expenses is normally shaped. A) Use the t-distribution. Correct answer B) Use the normal distribution. C) Cannot use the normal distribution or t-distribution. No Is the population normally shaped? Yes Do we know the population standard deviation? No

The sample size is less than 30. The population has a normal distribution. The population standard deviation is unknown. So we will use the t-distribution. Find the degrees of freedom? 24 2. 064

Find the margin of error

Interpret the confidence interval. We are 95% confident that the interval from 77. 29 to 85. 71 captures the actual mean of the population.

When 475 college seniors were surveyed, 125 said they had taken a college statistics course. Find a point estimate p, the population proportion of college seniors who had taken a college statistics course. Round to two decimal places.

In a survey of 2480 golfers, 15% said they were left-handed. The survey’s margin of error was 3%. Construct a confidence interval for the proportion of left-handed golfers. Is our population approximately distributed? 372 2108

In a survey of 2480 golfers, 15% said they were left-handed. The survey’s margin of error was 3%. Construct a confidence interval for the proportion of left-handed golfers. Since we are already given the margin of error, we do not need to find the degrees of freedom or the critical value, t.

Interpret the confidence interval. We are 95% confident that the interval from. 12 to. 18 captures the true proportion of the population.

In a survey of 10 golfers, 2 were found to be left-handed. Is it practical to construct the 90% confidence interval proportion, p? . Explain. 2 8 Since one of the conditions is not met, it is not practical to construct a confidence interval.

A pollster wishes to estimate the proportion of U. S. voters who favor capital punishment. How large a sample is needed to be 90% confident that the sample proportion will not differ from the true proportion by more than 4%?

A pollster wishes to estimate the proportion of U. S. voters who favor capital punishment. How large a sample is needed to be 90% confident that the sample proportion will not differ from the true proportion by more than 4%? We need a minimum of 625 voters.

A pollster wishes to estimate the proportion of U. S. voters who favor capital punishment. A random sample of 625 voters was selected and it was found that 125 voters favor capital punishment. Construct a 90% confidence interval of the true proportion of the population.