The mean age of bus drivers in Chicago is 54.5 years. What is the set of hypotheses that should be used to test the age of the bus drivers? O Ho : µ= 54. 4 H: μ> 54.4 Но : а Ha : 54. 4 54. 4 %3D O Ho : µ = 54. 4 Ha: µ+ 54. 4 Ho : = 54. 4 Ha: > 54.4

Why would the null hypothesis not be the mean lifespan of sexually inactive males (62.5)?

Is narcissism a more common personality trait today than it was a few decades ago?  It is known that the mean population score on the Narcissistic Personality Inventory (NPI) for students attending University of South Alabama around 20 years ago was μ= 15 (Twenge, 2010).  Interested in the narcissism levels of students in the year 2020, a researcher administers the NPI to a random sample of 25 University of Alabama sophomores this Spring term. The mean NPI score from the researcher’s sample of sophomores is M = 16.5, with s = 3.4. 1. Write the null and alternative hypotheses in symbols. Possible symbols for your answer: H0, H1, μ, M, σ. 2. Calculate the standard error.  3. Find the critical value for the test statistic, assuming alpha = .05 (Use largest [i.e., most conservative] value if exact value not given in the chart)  a) 2.064 b) 1.96 c) 1.98 d) 2.000

What does the subtraction of essentially the same amount of variability due to individual differences from both the numerator and denominator of the F ratio causes?

Find the proportion of N.d. observations with a zz-score between z_1 =z1​= -2.45 and z_2 =z2​= 2.71. Give your answer to four decimal places.

The mean potassium content of a popular sports drink is listed as 132 mg in a 32-oz bottle. Analysis of 24 bottles indicates a sample mean of 131.2 mg. (a) State the hypotheses for a two-tailed test of the claimed potassium content.. a. Ho: μ= 132 mg vs. H₁: b. He: ≤132 mg vs. H₁: c. He: ≥132 mg vs. H₁: Oa Ob Oc (b) Assuming a known standard deviation of 2.1 mg, calculate the z test statistic to test the manufacturer's claim. (Round your answer to 2 decimal places. A negative value should be indicated by a minus sign.) Test statistic 132 mg >132 mg <132 mg (c) At the 2 percent level of significance (a = .02) does the sample contradict the manufacturer's claim? Decision Rule: Reject Ho

a consumer analyst reports that the mean life of a certain type of alkaline battery is no more than 63 months. write the null and alternative hypotheses and ote which is the claim

The level of dissolved oxygen in a river indicates the water’s ability to support aquatic life. A research project is started to determine if the oxygen levels in the river have changed. A study done 10 years prior, showed that the mean oxygen levels in milligrams per liter for the river was 4.3. A.) Write appropriate hypotheses for this test B.) Water is collected samples from 15 randomly chosen locations along the river and measured the dissolved oxygen. Construct and interpret a 95% confidence interval for the mean dissolved oxygen level for this stream. The results are in milligrams per liter:  4.53 5.5 4.01 5.04 4.83 4.66 3.29 4.4 2.87 5.23 5.42 5.73 4.13 6.38 5.55 C.) Based on the sample from part B, conduct a significance test of these results using a 5% significance level. Give complete results of the test including a p-value and what the results means.

A trial study suggest that the standard deviation for the half life for the drug chlordiazepoxide is 8 hours. With 99% confidence, find the sample size needed to determine the mean half life for the drug with a margin of error of 3 hours.

Low‑density lipoprotein, or LDL, is the main source of cholesterol buildup and blockage in the arteries. This is why LDL is known as "bad cholesterol." LDL is measured in milligrams per deciliter of blood, or mg/dL. In a population of adults at risk for cardiovascular problems, the distribution of LDL levels is normal, with a mean of 123123 mg/dL and a standard deviation of 4141 mg/dL. If an individual's LDL is at least 11 standard deviation or more above the mean, he or she will be monitored carefully by a doctor. What percentage of individuals from this population will have LDL levels 11 or more standard deviations above the mean? Use the 6868–9595–99.799.7 rule. (Enter your exact answer as a whole number.)