HYPOTHESIS TEST # 1 Just Born Candy Co. makes an average of 12,300 boxes of Hot Tamales per day with stdey = 318 boxes.Julio is the company statistician and he claims the average number of boxes has changed, so he does a sampling of 121 days andfinds a sample mean = 12,256 boxes with stdev= 318 boxes. At alpha = 0.01, DO NOT USE THE P-VALUE METHOD-USE THE TRADITIONALMETHOD-do not use at test. AFTER PLACING THE CRITICAL VALUE(S) AND THE TEST VALUEON THE PROPER FIGURE, THE TEST VALUE IS :X. IN THE CRITICAL REGION(S)Y. IN THE NON CRITICAL REGION(S)_Z. EXACTLY ON THE CRITICAL VALUE(S)What is the decision ?A.REJECT HOВ.REJECT H1С.DO NOT REJECT HOD.DO NOT REJECT H1Е.NONE OF THESE ANSWERSState the conclusion about this test using the proper format. CHOOSE THE LETTERTHAT APPLIES.NOTE: THE DECISION IS NOT THE CONCLUSIONA. STATISTICIAN IS CORRECT, IT IS LOWERB. STATISTICIAN IS CORRECT, IT IS HIGHERC. STATISTICIAN IS CORRECT, IT HAS CHANGEDD. STATISTICIAN IS INCORRECT, IT IS NOT LOWERE. STATISTICIAN IS INCORRECT, IT IS NOT HIGHERF. STATISTICIAN IS INCORRECT, IT HAS NOT CHANGED

Let Given Data is Candy makes average boxes μ=12300 of Hot Tamales per day with Standard Deviation…

Answered: -USE THE TRADITIONAL

Math

Statistics

-USE THE TRADITIONAL

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

1st Edition

ISBN:9781680331141

Author:HOUGHTON MIFFLIN HARCOURT

Publisher:HOUGHTON MIFFLIN HARCOURT

Chapter11: Data Analysis And Displays

Section11.4: Two-ways Tables

Problem 12E

See similar textbooks

Similar questions

Population Genetics In the study of population genetics, an important measure of inbreeding is the proportion of homozygous genotypesthat is, instances in which the two alleles carried at a particular site on an individuals chromosomes are both the same. For population in which blood-related individual mate, them is a higher than expected frequency of homozygous individuals. Examples of such populations include endangered or rare species, selectively bred breeds, and isolated populations. in general. the frequency of homozygous children from mating of blood-related parents is greater than that for children from unrelated parents Measured over a large number of generations, the proportion of heterozygous genotypesthat is, nonhomozygous genotypeschanges by a constant factor 1 from generation to generation. The factor 1 is a number between 0 and 1. If 1=0.75, for example then the proportion of heterozygous individuals in the population decreases by 25 in each generation In this case, after 10 generations, the proportion of heterozygous individuals in the population decreases by 94.37, since 0.7510=0.0563, or 5.63. In other words, 94.37 of the population is homozygous. For specific types of matings, the proportion of heterozygous genotypes can be related to that of previous generations and is found from an equation. For mating between siblings 1 can be determined as the largest value of for which 2=12+14. This equation comes from carefully accounting for the genotypes for the present generation the 2 term in terms of those previous two generations represented by for the parents generation and by the constant term of the grandparents generation. a Find both solutions to the quadratic equation above and identify which is 1 use a horizontal span of 1 to 1 in this exercise and the following exercise. b After 5 generations, what proportion of the population will be homozygous? c After 20 generations, what proportion of the population will be homozygous?
UESTION 29 The average UQ student reports dedicating 8.5 hours per week to each 2-unit course. What is the approximate t obtained value for a group of 20 psychology students who study for a mean of 10 hours per week for each 2-unit course, where the estimated population variance is 9. Would this sample be considered significantly different from the population, applying an alpha level of 0.05? t = 0.67; the sample would be considered statistically significant t = 2.24; the sample would be considered statistically significant t = 2.24; the sample would NOT be considered statistically significant t = 0.67; the sample would NOT be considered statistically significant
Professional Golfers’ Earnings Two random samples of earnings of professional golfers were selected. One sample was taken from the Professional Golfers Association, and the other was taken from the Ladies Professional Golfers Association. At α=0.10, is there a difference in the means? The data are in thousands of dollars. Use the critical value method with tables. PGA 446 1147 1344 9188 5687 4910 8553 7573 375 LPGA 48 76 122 466 863 100 1876 2029 4364 2921 (a)State the hypotheses and identify the claim. H0: claim/ do not claim H1: claim/do not claim This hypothesis test one tailed/two tailed a (Choose one) test. (b)Find the critical value. Round the answer to at least two decimal places. Critical value is . Critical value is (c)Compute the test value. Round the standard deviations and F score values to at least two decimal places as needed. F= (d)Make the decision.…
Water Consumption The Old Farmer's Almanac stated that the average consumption of water per person per day was 123 gallons. To test the hypothesis that this figure may no longer be true, a researcher randomly selected 21 people and found that they used on average 118.1 gallons per day and =s5.3. At =α0.10 , is there enough evidence to say that the Old Farmer's Almanac figure might no longer be correct? Use the P-value method. Assume the population is approximately normally distributed.
Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester. In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 40 expectant mothers have mean weight increase of 15.8 pounds in the second trimester, with a standard deviation of 6.2 pounds. A hypothesis test is done to see if there is evidence that weight increase in the second trimester is greater than 14 pounds. Find the p-value for the hypothesis test. The p-value should be rounded to 4 decimal places.
Polychlorinated biphenyls (PCBs) are a group of synthetic oil-like chemicals that were at one time widely used as insulation in electrical equipment and were discharged into rivers. They were discovered to be a health hazard and were banned in the 1970s. Since then, much effort has gone into monitoring PCB concentrations in waterways. Assume that water samples are being drawn from a waterway in order to estimate the PCB concentration. a) Suppose that a random sample of size 80 has a sample mean of 1.69 ppb and a sample standard deviation of 0.25 ppb. Find a 95% confidence interval for the PCB concentration. b) Estimate the sample size needed so that a 95% confidence interval will specify the population mean to within ±0.02 ppb.
An experiment was conducted to test if the presence of a witness (no witness, mirror, adult present) has an effect on the number of candies that children will attempt to steal from a Halloween jar. Children were randomly assigned to one of the three conditions. The alpha level was .01. What statistical test should you use to evaluate the hypothesis above? one-way independent ANOVA multiple linear regression Pearson Product-Moment Correlation simple linear regression two-way independent ANOVA State the alternative hypothesis in notation form for the hypothesis above. cannot be stated in notation form p =/ 0 u1 = u2 = u3 B =/ 0 u1 =/ u2 =/ u3
A health officer is trying to study the malaria situation of Zambia. From the records of seasonal blood survey (SBS) results he came to understand that the proportion of people having malaria in Zambia was 3.8% in 2015. The size of the sample considered was 15,000. He also realized that during the year that followed (2016), blood samples were taken from 10,000 randomly selected persons. The result of the 2016 seasonal blood survey showed that 200 persons were positive for malaria. Help the Health officer in testing the hypothesis that the malaria situation of 2016 did not show any significant difference from that of 2015 (take the 5% level of significance).
: Researchers studies the effect of a fertilizer on radish sprout growth. They randomly selected some radish seeds to serve as controls, while others were planted in aluminum planters to which fertilizer sticks were added. Other conditions were held constant between the two groups. The following table shows a summary of the heights of plants (in cm) 2 weeks after germination Perform a two-sample (pooled) test to investigate whether the average population height of plants is significantly different between the two groups at 5% level.Step 1: H0:_____________________ and Ha: _____________________ Significance level α=_____________Step 2: List and verify necessary data conditions and…