Question

In: Statistics and Probability

The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data table....

The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data table. Use the data to​ (a) find the sample​ mean, (b) find the sample standard​ deviation, and​ (c) construct a​ 98% confidence interval for the population mean.

100,700 81,419 77,543 68,099 51,620 67,741 94,804 66,410 79,448 73,448 44,319 86,757 60,644 58,151 54,634 78,891

47,896 98,388 80,793 92,543 63,410 75,100 50,972 60,936 91,207 84,375 80,002 63,941 74,232 56,985 46,910 89,548

76,023 62,168 82,712

Solutions

Expert Solution

The table given below ,

X
100700 819155344.4
81419 87233545.21
77543 29854093.93
68099 15841275.61
51620 418575182
67741 18819198.37
94804 516420625.5
66410 32138808.19
79448 54300539.83
73448 1873859.832
44319 770623707.2
86757 215440454.9
60644 130761740.7
58151 193992248.2
54634 304331862.9
78891 46401845.37
47896 584822809.3
98388 692157693
80793 75931878.93
92543 418770793.9
63410 75153468.19
75100 9125776.392
50972 445510092.6
60936 124168900.5
91207 365876175.9
84375 151188910.9
80002 62772185.95
63941 66228834.37
74232 4634935.352
56985 227832156.7
46910 633484098.2
89548 305162117.8
76023 15554268.33
62168 98230101.43
82712 113058349.8

From table ,

a) The sample mean is given by ,

b) The sample standard deviation is given by ,

c) The degrees of freedom is given by ,

d.f.=n-1=35-1=34

The critical value is given by ,

; From excel "=TINV(0.02,34)"

Therefore , the 98% confidence interval for the population mean is given by ,

orchestra answered 2 years ago
Next > < Previous

Related Solutions

The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data table....
The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data table. Use the data to​ (a) find the sample​ mean, (b) find the sample standard​ deviation, and​ (c) construct a​ 98% confidence interval for the population mean. 100,095 81,395 77,851 68,184 51,784 68,158 94,436 66,446 75,556 78,621 74,185 44,588 86,744 60,998 57,664 55,388 78,844 61,904 47,480 98,368 80,618 93,023 63,344 75,122 51,049 60,873 82,029 91,267 84,094 79,834 63,872 73,989 57,207 46,919 88,912 The monthly incomes...
1. The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data...
1. The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data table. Use the data to​ (a) find the sample​ mean, (b) find the sample standard​ deviation, and​ (c) construct a​ 98% confidence interval for the population mean. 99 comma 562 80 comma 800 78 comma 478 67 comma 513 51 comma 609 68 comma 005 93 comma 538 65 comma 987 78 comma 937 73 comma 393 44 comma 374 86 comma 745 61...
V. The following data give the annual salaries (in thousand dollars) of 20 randomly selected health...
V. The following data give the annual salaries (in thousand dollars) of 20 randomly selected health care workers. 50 71 57 39 45 64 38 53 35 62 74 40 67 44 77 61 58 55 64 59 Prepare a box-and-whisker plot. Is the data set skewed in any direction? If yes, is it skewed to the right or to the left? Does this data set contain any outliers?
a) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.36 million dollars....
a) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.36 million dollars. Assuming a population standard deviation gross earnings of 0.48 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions). Confidence interval: ( , ). b) Which of the following is the correct interpretation for your answer in part (a)? A. We can be 99% confident that the mean gross earnings of all Rolling Stones concerts...
a) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.79 million dollars....
a) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.79 million dollars. Assuming a population standard deviation gross earnings of 0.47 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions). Confidence interval: ( __________________ , __________________ ).
The table below lists weights​ (carats) and prices​ (dollars) of randomly selected diamonds. Find the​ (a)...
The table below lists weights​ (carats) and prices​ (dollars) of randomly selected diamonds. Find the​ (a) explained​ variation, (b) unexplained​ variation, and​ (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear​ correlation, so it is reasonable to use the regression equation when making predictions. For the prediction​ interval, use a​ 95% confidence level with a diamond that weighs 0.8 carats. Weight 0.3 0.4 0.5 0.5 1.0 0.7 Price ​$517 ​$1163 ​$1350 ​$1410 ​$5672 ​$2278...
The table below lists weights​ (carats) and prices​ (dollars) of randomly selected diamonds. Find the​ (a)...
The table below lists weights​ (carats) and prices​ (dollars) of randomly selected diamonds. Find the​ (a) explained​ variation, (b) unexplained​ variation, and​ (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear​ correlation, so it is reasonable to use the regression equation when making predictions. For the prediction​ interval, use a​ 95% confidence level with a diamond that weighs 0.8 carats. Weight   Price 0.3   520 0.4   1171 0.5   1336 0.5   1425 1.0   5671 0.7   2278
The heights and weights of 15 randomly selected adult men are shown in the table. Use...
The heights and weights of 15 randomly selected adult men are shown in the table. Use α = .05 . Ht (in.) 73 70 73 72 68 72 68 70 68 67 72 70 65 74 70 Wt (lbs.) 185 155 195 164 145 185 170 150 180 175 175 147 140 210 200 State the conclusion in words. There is not sufficient sample evidence to support the claim of linear correlation. The sample data support the claim of linear...
1) For 31 randomly selected Rolling Stones concerts, the mean gross earnings is 2.61 million dollars....
1) For 31 randomly selected Rolling Stones concerts, the mean gross earnings is 2.61 million dollars. Assuming a population standard deviation gross earnings of 0.5 million dollars, obtain a 99% confidence interval (assume C-Level=0.99) for the mean gross earnings of all Rolling Stones concerts (in millions). Confidence interval: ___,___ 2) correct interpretation for part 1 answers? a. 99% chance that the mean gross earnings of all rolling stones concerts lies in the interval b. 99% confident that the mean gross...
The annual earnings of 14 randomly selected computer software engineers have a sample standard deviation of...
The annual earnings of 14 randomly selected computer software engineers have a sample standard deviation of $3630. Assume the sample is from a normally distributed population. Construct a confidence interval for the population variance σ2   the population standard deviation σ. Use a 99% level of confidence. Interpret the results. The waiting times​ (in minutes) of a random sample of 22 people at a bank have a sample standard deviation of 3.3 minutes. Construct of confidence interval for the population variance...
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT