A researcher begins with a known population-in this case, scores on a standardized test that are normally distributed with ? = 82.3 and ? = 15. The researcher suspects that special training in reading skills will produce an increased change in the scores for the individuals in the population. Because it is not feasible to administer the treatment (the special training) to everyone in the population, a sample of n=25 individuals is selected, and the treatment is given to this sample. Following treatment, the average score for this sample is x?=86. Is there evidence that the training has an effect on test scores? Perform a one-way directional test and complete the 5 step process. (?=.05) (4 points) H0: ?= (Example answer ##) Ha:?> (Example answer ##) Ztest= (Example answer #.##) p-value= (Example answer .####) Fail to reject or Reject the (Example answer Failed to reject or Reject)
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3 (H0: μ = 82.3) - Alternative Hypothesis (HA): The mean test score is greater than 82.3 (HA: μ > 82.3) ** Show more…
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1. A researcher begins with a known population – in this case, scores on a standardized test that are normally distributed with a mean of 65 and population standard deviation of 15. The researcher suspects that special training in reading skills will produce a change in the scores for the individuals in the population. Because it is not feasible to administer the treatment (the special training) to everyone in the population, a sample of 25 individuals is selected, and the treatment is given to this sample. Following treatment, the average score for this sample is 70. Is there evidence that the training has an effect on test scores? Use a significance level of .01. Also, find the 96% and 98% confidence interval of the true mean. Hint; use normal distribution table. Determine the p-values, and interpret its meaning.
Supreeta N.
We test a technique for changing reading comprehension. Without it, the population mean on a reading test is 220 with a standard deviation of 15. An sample of 25 participants has a mean of 211.55. Use a Z test and the 10 step process to determine if the technique changes reading comprehension scores. Use this as a sample to complete the rest. Step 1: Null Hypothesis - There is no difference in reading comprehension scores between participants who use the new technique and those who don't. Step 2: Research Hypothesis - There is a difference in the scores of participants who use the new technique for reading comprehension. Step 3: This test will be two-tailed Step 4: The level of significance is .05 Step 5: Use the Z test Step 6: N/A Step 7: Z crit = -1.96 or +1.96 Step 8: If Z obt is beyond +1.96 or -1.96, I must reject the null and accept the research hypothesis. Step 9: Z obt = -2.82 Step 10: There is sufficient evidence at the 5% level to support the claim that the alternative hypothesis represents the population.
David N.
In a study of babies born with very low birth weights, 275 children were given IQ tests at age 8 , and their scores approximated a normal distribution with $\mu=95.5$ and $\sigma=16.0$ (based on data from "Neurobchavioral Outcomes of School-age Children Born Extremely Low Birth Weight or Very Preterm," by Anderson ct al., Journal of the American Medical Association, Vol. 289, No. 24). Fifty of those children are to be randomly selected without replacement for a follow-up study. a. When considering the distribution of the mean IQ scores for samples of 50 children, should $\sigma_{\bar{x}}$ be corrected by using the finite population correction factor? Why or why not? What is the value of $\sigma_{\bar{z}} ?$ b. Find the probability that the mean IQ score of the follow-up sample is between 95 and $105 .$
Normal Probability Distributions
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