Question

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8 cases per year. find a 90%, 95%, and 98% confidence interval for the population mean annual number of reported larceny cases in such communities. compare the lengths of the confidence intervals. as the confidence levels increase, do the confidence intervals increase in length?

Answer 1

When calculating the 90%, 95%, and 98% confidence intervals for the population mean annual number of reported larceny cases, the intervals widen as the confidence level increases due to the higher Z-values used in the calculation, reflecting higher certainty that the interval contains the true mean.

Step-by-step explanation:

To find the confidence intervals (CIs) for the population mean number of reported larceny cases in the twenty-two communities, we use the formula for a confidence interval when the population standard deviation (σ) is known:

CI = μ ± Z*(σ/√n)

where:

• μ is the sample mean,
• Z* is the Z-value from the standard normal distribution for the required confidence level,
• σ is the population standard deviation, and
• n is the sample size.

Here, μ = 123.6, σ = 36.8, and n = 22.

The Z-values for 90%, 95%, and 98% confidence levels can be found using a Z-table or standard software/tools: approximately 1.645, 1.96, and 2.33 respectively.

Using these values, the confidence intervals can be calculated:

• 90% CI: 123.6 ± 1.645*(36.8/√22)
• 95% CI: 123.6 ± 1.96*(36.8/√22)
• 98% CI: 123.6 ± 2.33*(36.8/√22)

As the confidence level increases, the Z-value increases, which in turn increases the margin of error and the length of the confidence interval. Thus, a higher confidence level means a broader range to estimate the population mean, indicating we are more confident that the range contains the true mean.

Answer 2

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval =
`m-z *(psd)/( √(n) )`
Using the values, we get:
Lower end of confidence interval=
`123.6-1.645* (36.8)/( √(22))=110.69`

Upper end of confidence interval =
`m+z *(psd)/( √(n) )`
Using the values, we get:
Upper end of confidence interval=
`123.6+1.645* (36.8)/( √(22))=136.51`

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval =
`m-z *(psd)/( √(n) )`
Using the values, we get:
Lower end of confidence interval=
`123.6-1.96* (36.8)/( √(22))=108.22`

Upper end of confidence interval =
`m+z *(psd)/( √(n) )`
Using the values, we get:
Upper end of confidence interval=
`123.6+1.96* (36.8)/( √(22))=138.98`

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval =
`m-z *(psd)/( √(n) )`
Using the values, we get:
Lower end of confidence interval=
`123.6-2.327* (36.8)/( √(22))=105.34`
Upper end of confidence interval =
`m+z *(psd)/( √(n) )`
Using the values, we get:
Upper end of confidence interval=
`123.6+2.327* (36.8)/( √(22))=141.86`

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.