You are asked to determine the estimated number of car accidents per month at a specific intersection. Your report needs to be on your supervisor's desk the next morning. Since it is an estimate, your supervisor asked for an estimate with 98% confidence. In response to this assignment, you randomly selected a sample of 16 months over the last few years and compiled data on the number of car accidents that occurred during each of these months. The mean number of car accidents in these months was 42 per month with a sample standard deviation of 9 per month. What estimate will your report indicate

The 98% confidence interval for the mean number of car accidents per month at a specific intersection is between 18.58 and 65.42 accidents per month.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.602

The margin of error is:

M = T*s = 2.602*9 = 23.42

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 23.42 = 18.58 accidents per month.

The upper end of the interval is the sample mean added to M. So it is 42 + 23.42 = 65.42 accidents per month.

The 98% confidence interval for the mean number of car accidents per month at a specific intersection is between 18.58 and 65.42 accidents per month.

which expression can be used to convert 80 us dollars (usd) to australian dollars (aud)?

1 usd = 1.0343 aud

1 aud = 0.9668 usd

step-by-step explanation:

sorry just need the points

hope this .

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