The test statistic for a left-tailed test is Z = -2.65. Determine the P-value. 0.0080 0.0040 0.9960 0.9920
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- Test the claim that the mean GPA of night students is greater than the mean GPA of day students at the 0.025 significance level.The distribution of heights in a population of women is approximately normal. Sixteen percent of the women have heights less than 62 inches. About 97.5% of the women have heights less than 71 inches. Use the empirical rule to estimate the mean and standard deviation of the heights in this population. Mean: K inches Standard Deviation: inchesFind the p-value for the z-test. (Round your answer to four decimal places.) a left-tailed test with observed z = -1.83 In USE SALT p-value = Determine the significance of the results. O The results are highly significant. The results are statistically significant. The results are only tending toward statistical significance. The results are not statistically significant.
- A test of reading skills is administered to 12th grade students. The distribution of scores is normal with mean 288 and standard deviation 38. Using the 68-95-99.7 rule, fill in the blanks in the statement below: 95% of reading scores will be between (blank) and (blank)Pls help ASAP. Pls show all work.The chi-square statistic for these data is x = 8.134. Which of the following intervals contains the P- value for this test? Color preference Pink Blue Yellow Green Total Female 48 24 18 30 120 Gender Male 55 34 18 116 Total 103 58 27 48 236 0.025 0.1
- 3K A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. e this 2 Z a. Use a 0.05 significance level to test the claim that men have a higher mean body temperature than women. What are the null and alternative hypotheses? OA Ho: Hy #₂ H₂: Hy H₂ F2 淋 X command дв 3 80 F3 E D с $ 4 R F % 5 V T G 6 B MacBook Air Y H & 7 N F7 U J * 8 M OB. Ho: ₁2₂ H₁: Hy t : - ; H n X S option Clear all F11 { [ Men H₁ 11 97.64°F 0.77 F ? 1 + = 11 I Check answer 4) F12 Women H₂ 59 97.45°F 0.63°F } ] delete ▼ return shift drewK Assume that adults have IQ scores that are normally distributed with a mean of 101.5 and a standard deviation 20.1. Find the first quartile Q₁, which is the IQ score separating the bottom 25% from the top 75%. (Hint: Draw a graph.) The first quartile is (Type an integer or decimal rounded to one decimal place as needed.)
- K A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal A. Ho H1 H2 H₁ H₁ H₂ ỌC. Hồ Hi-12 H₁: Hy > H₂ 0 в. Но 141=12 H₁ H₁ H₂ O D. Ho 11 #₂2 H₁ H1 H₂ The test statistic, t, is (Round to two decimal places as needed) The P-value is (Round to three decimal places as needed.) State the conclusion for the test. Proctored Nonproctored H 144 n 34 CIXS 77.58 11.08 12 31 81.91 21.24 A. Reject Ho There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. OB. Fail to reject Ho. There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. C. Fail to reject Ho. There is not sufficient evidence to support the…11 Scores on a dental anxiety scale range from zero (no anxiety) to 20 (extreme anxiety). The scores are normally distributed distributed with a mean of 11 and a standard deviation of four. Find the Z score for the given score on this dental anxiety scale. 5 z5= ____ (Type an integer or a decimal)You are performing a two-tailed test with test statistic z=−1.793, find the p-value accurate to 4 decimal places.p-value = ________