Question

1.) A randomly selected sample of 70 casino patrons has an average loss of $300 with a sample standard deviation of $100. Use this information to complete the following:

a) Does the information above provide an estimate of a population mean or a population proportion?

b) Provide a point estimate for the population mean loss of all casino patrons.

c) Construct and interpret a 90% confidence interval for the parameter you selected in part (a).

d) What is the margin of error for this confidence interval estimate?

Answer #1

a)

population mean (since it is the estimate of average loss suffered by all casino patrons )

b)

point estimate =sample mean =300

c)

sample mean 'x̄= | 300.000 | |

sample size n= | 70 | |

std deviation s= | 100.000 | |

std error ='sx=s/√n=100/√70= | 11.9523 |

for 90% CI; and 69 df, value of t= | 1.667 |
from excel: t.inv(0.95,69) |
||

margin of error E=t*std error = | 19.93 | |||

lower bound=sample mean-E = | 280.073 | |||

Upper bound=sample mean+E = | 319.927 | |||

from above
90% confidence interval for population mean
=(280.07<μ<319.93) |

above interval gives 90% confidence to contain true value
of population mean |

d)

**margin of error for this confidence interval estimate
=19.93**

A random sample of 50 students has a mean GPA of 2.55. It is
known that the population standard deviation for the GPA of all
students is 1.1. Use this information to complete the
following:
a) Does the information above provide an estimate of a
population mean or a population proportion?
b) Provide a point estimate for the population mean GPA of all
students.
c) Construct and interpret a 95% confidence interval for the
parameter you selected in part (a)....

Randomly selected students participated in an experiment to test
their ability to determine when one minute (or sixty seconds) has
passed. Forty students yielded a sample mean of 59.7 seconds.
Assuming that sigma =10.1 seconds, construct and interpret a 90%
confidence interval estimate of the population mean of all
students.
What is the 90% confidence interval for the population mean
μ?

2. You randomly select and measure the lengths of 34
bolts. The sample mean length is 1.24 inches and the standard
deviation is 0.3 inches. Construct a 94% interval of confidence of
the mean length of all bolts.
Find a point estimate for the mean length of all
bolts
Verify that the conditions are met to construct an
interval of confidence for the mean length of all bolts
Calculate the critical Z value to construct a 94%
confidence interval (round...

Use the sample data and confidence level given below to complete
parts (a) through (d).
In a study of cell phone use and brain hemispheric dominance,
an Internet survey was e-mailed to 2480 subjects randomly selected
from an online group involved with ears. 1030 surveys were
returned. Construct a 90% confidence interval for the proportion
of returned surveys.
a) Find the best point estimate of the population
proportion p.
(Round to three decimal places as needed.)
b) Identify the value...

A research agency conducted a survey on randomly selected 2000 Malaysian adults aged 18 and over regarding weight issue. These adults were asked if they want to lose weight and 63% said yes.
i) Construct a 97% confidence interval for the corresponding population proportion.
ii) Find the minimum sample size if the research agency wants to be sure that the margin of error is within 0.04 of the population proportion for a 97% confidence interval. The preliminary estimate for the...

The
annual earnings of 13 randomly selected computer software engineers
have a sample standard deviation of $ 3660. Assume the sample is
from a normally distributed population. Construct a confidence
interval for the population variance sigma squared and the
population standard deviation sigma. Use a 90 % level of
confidence. Interpret the results.

The data set represents the scores of 12 randomly selected
students on the SAT Physics Subject Test. Assume the population
test scores are normally distributed and the population standard
deviation is 103. (Adapted from The College Board) 670 740 630 620
730 650 720 620 640 500 670 760
(a) Find the point estimate of the population mean.
(b) Construct a 90% confidence interval for the population mean.
Interpret the results.
(c) Determine the minimum sample size required to be...

A researcher measured the body temperatures of a randomly
selected group of adults. He wishes to estimate the average
temperature among the adult population. Summaries of the data he
collected are presented in the table below. Complete parts? (a)
through? (d) below.
Summary
Count
Mean
Median
MidRange
StdDev
Range
IntQRange
tempeture
43
98.457
98.000
98.600
0.8847
2.800
1.050 ?
a) Would a 90?% confidence interval be wider or narrower than
the 98?% confidence? interval? Explain. Choose the correct...

Consider: Sugar content in a particular brand of cranberry
sauce. A random sample was selected in order to estimate
the true sugar content.
Sample results:
Sample size: 76
Sample Mean: 24.7 g
Sample standard deviation: 1.3 g.
a. Provide the 90% Confidence Interval for the
Mean amount of sugar per serving.
b. The sample result of ?̅= 24.7 ? can be used as a point
estimate for the mean amount of sugar per serving. At the 90%
Confidence Level, what is...

Fireworks. Last summer, Survey USA published results of a survey
stating that 281 of 531 randomly sampled Kansas residents planned
to set off fireworks on July 4th. Round all results to 4 decimal
places. 1. Calculate the point estimate for the proportion of
Kansas residents that planned to set off fireworks on July 4th 2.
Calculate the standard error for the point estimate you calculated
in part 1. 3. Calculate the margin of error for a 90 % confidence
interval...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 11 minutes ago

asked 16 minutes ago

asked 18 minutes ago

asked 23 minutes ago

asked 32 minutes ago

asked 36 minutes ago

asked 36 minutes ago

asked 37 minutes ago

asked 47 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago