please show me the work i dont know how to get these answers. Thank you! 1.Using
ID: 3155651 • Letter: P
Question
please show me the work i dont know how to get these answers. Thank you!
1.Using the t-tables, software, or a calculator, estimate the critical value of t for the given confidence interval and degrees of freedom.
A 90% confidence interval from a sample of size 20. A) 1.725 B) 1.734 C) 1.729 D) 2.093 E) 1.645
2.Use the t-tables, software, or a calculator to estimate the indicated P-value.
24) P-value for t 2.13 for a sample of size 31 A) 0.9793 B) 0.9653 C) 0.0207 D) 0.9480 E) 0.9913
25) P-value for t 2.13 for a sample of size 31 A) 0.0087 B) 0.0520 C) 0.0347 D) 0.9793 E) 0.0207
26) P-value for t 1.44 with 45 degrees of freedom A) 0.8431 B) 0.1569 C) 0.9216 D) 0.0785 E) 0.8914
3.Construct the requested confidence interval from the supplied information.
A researcher wants to estimate the mean cholesterol level of people in his city.A random sample of 21 people yields a mean cholesterol level of 224 and a standard deviation of 12. Construct a 95% confidence interval. A) (219.693, 228.307) B) (218.538, 229.462) C) (219.598, 228.402) D) (223.014, 224.986) E) (214.967, 233.033)
4.Determine the margin of error in estimating the population parameter.
Based on a sample of 39 randomly selected years, a 90% confidence interval for the mean annual precipitation in one city is from 42.8 inches to 45.2 inches. A) 0.10 inches B) 0.32 inches C) 2.4 inches D) 1.2 inches E) Not enough information is given.
Explanation / Answer
A 90% confidence interval from a sample of size 20. A) 1.725 B) 1.734 C) 1.729 D) 2.093 E) 1.645
df = 19 alpha / 2 = 0.05 t = 1.729
2.Use the t-tables, software, or a calculator to estimate the indicated P-value.
24) P-value for t 2.13 for a sample of size 31 A) 0.9793 B) 0.9653 C) 0.0207 D) 0.9480 E) 0.9913
p value is 1 - 0.0207 = 0.9793
25) P-value for t 2.13 for a sample of size 31 A) 0.0087 B) 0.0520 C) 0.0347 D) 0.9793 E) 0.0207
p value : 0.0207
3.Construct the requested confidence interval from the supplied information.
A researcher wants to estimate the mean cholesterol level of people in his city.A random sample of 21 people yields a mean cholesterol level of 224 and a standard deviation of 12. Construct a 95% confidence interval. A) (219.693, 228.307) B) (218.538, 229.462) C) (219.598, 228.402) D) (223.014, 224.986) E) (214.967, 233.033)
alpha / 2 = 0.025 df = 20 t = 2.086
I: 224 +/- 2.086 * 12 / srqt 21
224 +/- 5.46
218.54 < miu < 229.46
4.Determine the margin of error in estimating the population parameter.
Based on a sample of 39 randomly selected years, a 90% confidence interval for the mean annual precipitation in one city is from 42.8 inches to 45.2 inches. A) 0.10 inches B) 0.32 inches C) 2.4 inches D) 1.2 inches E) Not enough information is given.
alpha / 2 = 0.05 df = 38 t = 1.684
margin of error : not enough information is given , we need SD for the margin of error!
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