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In: Math
A simple random sample of checks were categorized based on the number of cents on the...
A simple random sample of checks were categorized based on the number of cents on the written check and recorded below. Cents Category 0¢-24¢ 25¢-49¢ 50¢-74¢ 75¢-99¢ Frequency 58 37 28 17 Use the p-value method and a 5% significance level to test the claim that 50% of the check population falls into the 0¢-24¢ category, 20% of the check population falls into the 25¢-49¢ category, 16% of the check population falls into the 50¢-74¢ category, and 14% of the check population falls into the 75¢-99¢ category. Calculate the expected value for the 50¢-74¢ category (round to the nearest tenth).
Solutions
milcah answered 7 months agoRelated Solutions
A simple random sample of checks were categorized based on the number of cents on the...
A simple random sample of checks were categorized based on the
number of cents on the written check and recorded below.
Cents Category 0¢-24¢ 25¢-49¢ 50¢-74¢ 75¢-99¢
Frequency 58 37 28 17
Use the critical value method and a 1% significance level to
test the claim that the frequencies of the cents categories of
checks fit the uniform distribution.
Calculate the expected frequency of the 25¢-49¢ category.
The number of successes and the sample size are given for a simple random sample from...
The number of successes and the sample size are given
for a simple random sample from a population. Use the
one-proportion z-interval procedure to find the required confidence
interval.
n = 76, x = 31; 98% level
0.298 to 0.518
0.276 to 0.540
0.297 to 0.519
0.277 to 0.539
Use the one-proportion z-interval procedure to find the
required confidence interval.
A researcher wishes to estimate the proportion of adults in the
city of Darby who are vegetarian. In a random...
The heights were recorded for a Simple Random Sample of 270 junior. The mean of this sample...
The heights were recorded for a Simple Random Sample of 270
junior. The mean of this sample was 66.5
inches. The heights are known to be Normally Distributed
with a population standard deviation of 5.1 inches. (You do
not need a data set for this). Test the claim that the
mean height of Juniors has increased from 65.7 at a 0.01
significance level. (Use MINITAB to do the hypothesis test
and copy and paste the output of the hypothesis test here (0.5pts)....
In a simple random sample of 100 households, the sample mean number of personal computers was...
In a simple random sample of 100 households, the sample mean
number of personal computers was 1.32. Assume the population
standard deviation is 0.41. Construct a 95% confidence interval for
the mean number of personal computers.
(a) (1.24, 1.40)
(b) (1.25, 1.39)
(c) (0.15, 0.67)
(d) (0.19, 0.63)
A simple random sample was conducted of 1405 Canadian adults. They were asked whether they were...
A simple random sample was conducted of 1405 Canadian adults.
They were asked whether they were in favor of tighter enforcement
of tobacco laws. Of the 1405 adults, 1060 responded yes. Obtain a
95% confidence interval for the
proportion of Canadian adults who are in favor of
tighter enforcement of tobacco laws.
Of 15, 344 Virginians tested for COVID-19 in March of 2020,
1,484 had positive or presumptive positive results for the virus.
Assuming the other 49 states had results...
The sample data below have been collected based on a simple random sample from a normally...
The sample data below have been collected based on a simple
random sample from a normally distributed population. Complete
parts a and b. 5 4 0 5 7 6 9 0 8 4 a. Compute a 98% confidence
interval estimate for the population mean. The 98% confidence
interval for the population mean is from to . (Round to two decimal
places as needed. Use ascending order.) b. Show what the impact
would be if the confidence level is increased to...
1) In a simple random sample of size 70, there were 35 individuals in the category...
1) In a simple random sample of size 70, there were 35
individuals in the category of interest. It is desired to test H0 :
p = 0.52 versus H1 : p < 0.52. Compute the test statistic z. 1)
A) 2.37 B) 0.50 C) 0.06 D) -0.33
Suppose a simple random sample of athletes in the NBA heights is taken. There were 28...
Suppose a simple random sample of athletes in the NBA heights is
taken. There were 28 athletes in the sample with a mean height of
78.4 inches and standard deviation of 2 inches. It has been
confirmed through statistical analysis that NBA player heights
follows a normal distribution.
a. what parameter are we estimating?
b. Explain the requirements as they relate to the problem
c.What is the point estimate of the parameter?
d.Input the margin of error for a 95%...
The average number of pages for a simple random sample of 40 physics textbooks is 435....
The average number of pages for a simple random sample of 40
physics textbooks is 435. The average number of pages for a simple
random sample of 30 mathematics textbooks is 410. Assume that all
page length for each types of textbooks is normally distributed.
The standard deviation of page length for all physics textbooks is
known to be 55, and the standard deviation of page length for all
mathematics textbooks is known to be 55. Part One: Assuming that...
In a simple random sample of 150 households, the mean number of personal computers was 1.32....
In a simple random sample of 150 households, the mean number of
personal computers was 1.32. Assume the population standard
deviation is σ = 0.41.
a. Construct a 95% confidence interval for the mean number of
personal computers.
b. If the sample size were 100 rather than 150, would the margin
of error be larger or smaller than the result in part (a)?
c. If the confidence level were 98% rather than 95%, would the
margin of error be larger...
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