(a) in 23 to 35 days chicks (b) in 26 to 32 days chicks (c) in 29 days or fewer chicks (d) in 20 to 38 days chicks

Chapter 5, Section 2, Exercise 043 Hearing Loss in TeenagersA recent study1 found that, of the 1771 participants aged 12 to 19 in the National Health and Nutrition Examination Survey, 19.5% had some hearing loss (defined as a loss of 15 decibels in at least one ear). This is a dramatic increase from a decade ago. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error for the proportion is SE=0.009.Find a 95% confidence interval for the proportion of teenagers with some hearing loss.Round your answers to two decimal places.The 95% confidence interval is to .

A quiz related to digital knowledge was conducted. The quiz had 10 questions and covered topics such as the purpose of browser cookies, phishing scams, and privacy policies. The survey was given to 50 people. The mean score is 4.1 with a standard deviation of 2.6. We want to know if the data provide evidence that the mean score in the population is lower than 5. We will use alpha = 0.05 to make our decision. part a. Is this a one-sided or two-sided test? part b. What is the standard error for this problem? part c. How many degrees of freedom do we have for this problem? part d. What is the test statistic, t, for this problem? Based on your test statistic and degrees of freedom, the p-value will be between what two values? (If you cannot find the exact df in the table, round down to be safe. So if you have 38 df, round down to 30df.) Group of answer choices The p-value is between .1 and .05 The p-value is between .05 and .025 The p-value is between .025 and .01 The p-value…

QUESTION 22 Use the 68-95-99.7 rule to solve the problem. The amount of Jen's monthly phone bill is normally distributed with a mean of $56 and a standard deviation of $12. Fill in the blanks. and $__. 68% of her phone bills are between $ 32, 56 56, 68 32, 80 44, 68 0000

Consider the data from the Anthropology 105 class.  The mean for women is 64.33 in and the standard deviation is 2.64 in. The average height of men in the US is approximately 5ft 10in. What proportion of women represented here are shorter than the average man?

Step 1: Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 70% of vehicles on the roads in the Minneapolis-St. Paul metropolitan area are occupied by just the driver. You choose 84 vehicles at random. What are the mean and standard deviation of the proportion of vehicles in your sample that carry only one person? (Fill in the blanks below.) The mean is (1) . (Give your answer to 1 decimal place.) (2) The standard deviation is (Give your answer to 2 decimal places.) eBook Answer 1 Answer 2

Section 9.2 Question #7 Listed in the data table are IQ scores for a random sample of subjects with medium lead levels in their blood. Also listed are statistics from a study done of IQ scores for a random sample of subjects with high lead levels. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below.   View the data table of IQ scores. Medium Lead Level     High Lead Level   72   n2 = 11   96   92   _ x2 = 91.815   85   90   s2 = 10.365   97   83   92   104   111   91   a. Use a 0.05 significance level to test the claim that the mean IQ scores for subjects with medium lead levels is higher than the mean for subjects with high lead levels.   What are the null and alternative​ hypotheses? Assume that population 1 consists of subjects with medium lead levels and population…

Question 30: give the test statistic

Question 10. (BALANCE SHEET IS MUST) A saleswoman sold two items with a sample mean price of $2 and a sample variance of 2 dollars-squared. Find the prices of the two products. а. 3 O b. 1.25 O c. 2.75 O d. 1