In order to evaluate the effectiveness of a new type of plant food that was developed for tomatoes, a study was conducted in which a random sample of n = 76 plants received a certain amount of this new type of plant food each week for 14 weeks. The variable of interest is the number of tomatoes produced by each plant in the sample. The table below reports the descriptive statistics for this study: Descriptive statistics for the tomato food study Variable n sample mean sample standard deviation standard error number of tomatoes 76 33.40 8.60 0.986 Assuming the population is normally distributed, the investigators would like to construct a 98% confidence interval for the average number of tomatoes that all plants of this variety can produce when fed this supplement like this. a) The margin of error is: 2.426 Correct: Your answer is correct. (3 decimals) b) The corresponding 98% confidence interval for the true population mean is: Lower Limit: Incorrect: Your answer is incorrect. (3 decimals) to Upper Limit: Incorrect: Your answer is incorrect. (3 decimals) c) What would we conclude at α = 0.02 for the hypothesis test H0: μ = 37.845 vs. Ha: μ ≠ 37.845? We do not have enough evidence to conclude the true mean is 37.845. We have sufficient evidence to conclude that the true mean is different from 37.845. We have insufficient evidence to conclude the true mean is different from 37.845. We have enough evidence to conclude that the true mean is 37.845. We do not have enough evidence to conclude the true mean is 33.40.
Given X bar = 33.40, Sd = 8.60, SE = 0.986 and n = 76
C = 98%, Alpha = 1-c = 0.02 and df = n-1 = 75
tc = 2.3771 (Use t table)
a) margin of Error ME = tc*SE = 2.3771*0.986 = 2.3438
b)
| CI | ||
| tc | 2.377102 | T.INV.2T(D1,D2) |
| Upper | 35.745 | X bar + tc*(S/SQRT(n)) |
| Lower | 31.055 | X bar - tc*(S/SQRT(n)) |
98% CI = (31.055, 35.745)
c)
Hypothesis
H0: μ = 37.845
Ha: μ ≠ 37.845
| Test : | ||
| t | -4.506 | (X bar-μ )/(S/SQRT(n)) |
| P value | 2.389E-05 | T.DIST.2T(-ts,df) |
P value < 0.02,Rejrct H0
We have sufficient evidence to conclude that the true mean is different from 37.845
In order to evaluate the effectiveness of a new type of plant food that was developed...
In order to evaluate the effectiveness of a new type of plant food that was developed for tomatoes, a study was conducted in which a random sample of n = 72 plants received a certain amount of this new type of plant food each week for 14 weeks. The variable of interest is the number of tomatoes produced by each plant in the sample. The table below reports the descriptive statistics for this study: Variable Descriptive statistics for the tomato...
A research team is interested in the effectiveness of hypnosis in reducing pain. The responses from 8 randomly selected patients before and after hypnosis are recorded in the table below (higher values indicate more pain). Construct a 90% confidence interval for the true mean difference in pain after hypnosis. Perceived pain levels 'Pre' and 'Post' hypnosis for 8 subjects Pre 11.3 8.1 8.4 11.1 13.4 15.1 10.6 10.1 Post 9.9 9.1 3.1 8.5 7.0 9.9 10.9 7.7 Difference a) Fill...
According to a food website, the mean consumption of popcorn annually by Americans is 57 quarts. The marketing division of the food website unleashes an aggressive campaign designed to get Americans to consume even more popcorn. Complete parts (a) through (c) below. (a) Determine the null and alternative hypotheses that would be used to test the effectiveness of the marketing campaign. H0: ▼ p σ μ ▼ ≠ = _______? H1: ▼ σ p μ ▼ ≠ > < _______?...
A fast food restaurant recently added a new sandwich to its menu. To estimate the popularity of this sandwich, a random sample of 43 customers who ordered the sandwich were surveyed. Each of these customers were asked to rate this sandwich on a scale of 1 to 100. The manager used the resulting data to calculate a 95% confidence interval for the mean rating of the new sandwich. 95% confidence interval: (57.65, 66.84) (a) Identify the parts of this confidence...
Question 1 A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours. A similar study conducted a year earlier estimated that μ, the mean number of weekly hours that U.S. adults use computers at home,...
7. A study was conducted to investigate the effectiveness of a new drug for treating Stage 4 AIDS patients. A group of AIDS patients was randomly divided into two groups. One group received the new drug; the other group received a placebo. The difference in mean subsequent survival (those with drugs - those without drugs) was found to be 1.04 years and a 95% confidence interval was found to be 1.04 ± 2.37 years. Based upon this information: Select one...
8 plant wants to determine whether the mean manager performance ratings ed over the past month. The table shows the plant's perfomance ratings for 3. A manufacturing want the same six managers last month an d this month. 6 78 89 5 81 92 4 76 86 Manager Number1 Rating (last month) 85 Rating (this month) 88 2 96 85 70 89 Sample standard deviation Summary Statistics Rating (last month) Rating (this month) Difference Mean 81.00 88.17 8.90 2.48 10.24...
A electronics manufacturer has developed a new type of remote control button that is designed to operate longer before failing to work consistently. A random sample of 29 of the new buttons is selected and each is tested in continuous operation until it fails to work consistently. The resulting lifetimes are found to have a sample mean of x¯x¯ = 1261.3 hours and a sample standard deviation of s = 119.4. Independent tests reveal that the mean lifetime of the...
1. A research company surveyed 650 adults in the US to estimate what percent of the country were afraid of their identity being stolen. From the sample, 408 indicated that they were afraid of their identity being stolen. Find the 92% confidence interval for the proportion of all adults in the US who are afraid of their identity being stolen. Round to 4 decimal points. 2. A laboratory tested 90 chicken eggs and found that the mean amount of cholesterol...
1. You measure 42 textbooks' weights, and find they have a mean weight of 47 ounces. Assume the population standard deviation is 3.5 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places 2.If n=16, ¯xx¯(x-bar)=43, and s=13, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place. 3.SAT scores are...