A random sample of 20 recent weddings in a country yielded a mean wedding cost of...
A random sample of 20 recent weddings in a country yielded a mean wedding cost of $ 26,323.26. Assume that recent wedding costs in this country are normally distributed with a standard deviation of $8400. Complete parts (a) through (c) below. a. Determine a 95% confidence interval for the mean cost, mu, of all recent weddings in this country.
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A random sample of 20 recent weddings in a country yielded a
mean wedding cost of...
A random sample of 20 recent weddings in a country yielded a mean wedding cost of...
A random sample of 20 recent weddings in a country yielded a mean wedding cost of $26,387.22. Assume that recent wedding costs in this country are normally distributed with a standard deviation of $8400. Complete parts (a) through (c) below. a. Determine a 95% confidence interval for the mean cost, u, of all recent weddings in this country. The 95% confidence interval is from $to $ (Round to the nearest cent as needed.) b. Interpret your result in part (a)....
A random sample of 20 recent weddings in a country yielded a mean wedding cost of...
A random sample of 20 recent weddings in a country yielded a mean wedding cost of $26,327.04. Assume that recent wedding costs in this country are normally distributed with a standard deviation of $7700. Complete parts (a) through (c) below. a. Determine a 95% confidence interval for the mean cost, μ, of all recent weddings in this country. The 95% confidence interval is from $___ to $___. b) Interpret your result in part (a). Choose the correct answer below. A.We...
a randomnsample of 20 recent weddings in a country yeilded a mean wedding cost of $26,368.73....
a
randomnsample of 20 recent weddings in a country yeilded a mean
wedding cost of $26,368.73. assume that recent wedding costs in
this country are normally distributed with a standard deviation of
$8100. complete parts A-C below
This Question: 1 pt 12 of 12 (5 complete) Arandom sample of 20 ecent weddings in a ceuntry yelded a mean wedding oost of $26 3873 Assume that recent wedding costs in this country ae nomally dstributed wth a standand deviation of 500...
Many couples believe that it is getting too expensive to host an "average" wedding in a...
Many couples believe that it is getting too expensive to host an "average" wedding in a certain country. According to a wedding cost website, the average cost of a wedding in the country in 2009 was $24,066. Recently, in a random sample of 40 weddings in the country it was found that the average cost of a wedding was $23,224, with a standard deviation of $2,903. On the basis of this, a 95% confidence interval for the mean cost of...
In a random sample of six mobile devices, the mean repair cost was $80.00 and the...
In a random sample of six mobile devices, the mean repair cost was $80.00 and the standard deviation was $12.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 95% confidence interval for the population mean. Interpret the results. The 95% confidence interval for the population mean mu is
A test on a random sample of 50 water balloons yielded a sample average weight of...
A test on a random sample of 50 water balloons yielded a sample average weight of 1.2 pounds. Prior studies have shown that the population standard deviation is 0.2 pounds. Assume that water balloon weight is normally distributed. Construct a 95% confidence interval of the population mean?
A test on a random sample of 26 tricornes (i.e., pirate hats) yielded a sample average...
A test on a random sample of 26 tricornes (i.e., pirate hats) yielded a sample average length of 23 inches and a sample standard deviation of 1.5 inches. Assume that tricorne length is normally distributed. How many total samples are needed to construct a 95% confidence interval that differs from the sample mean by at most 0.5 inches?
A random sample of 18 venture capital investments in the fiber optics business sector yielded a sample mean of $6.33 million.
A random sample of 18 venture capital investments in the fiber optics business sector yielded a sample mean of $6.33 million. What is the critical z-score value, zc, for an 80% confidence interval? Provide a screenshot of your answer.Use your answer in (1.) to determine the margin of error for an 80% confidence interval for the mean amount of all venture-capital investments in the fiber optics business sector. Assume that the population is normally distributed and the population standard deviation is...
A random sample of 10 venture-capital investments in the fiber optics business sector yielded the following...
A random sample of 10 venture-capital investments in the fiber optics business sector yielded the following data, in millions of dollars. Determine a 95% confidence interval for the mean amount, mu, of all venture-capital investments in the fiber optics business sector. Assume that the population standard deviation is $2.43 million. (Note: The sum of the data is $62.59 million.) The confidence interval for mu is from $ _____million to $______million. (Round to two decimal places as needed.)
In a random sample of six microwave ovens, the mean repair cost was $65.00 and the...
In a random sample of six microwave ovens, the mean repair cost was $65.00 and the standard deviation was $12.50. Assume the population is normally distributed and use a t-distribution to construct a 90% confidence interval for the population mean mu. What is the margin of error of mu? Interpret the results. The 90% confidence interval for the population mean mu is ( nothing, nothing). (Round to two decimal places as needed.) The margin of error is nothing. (Round to...
A random sample of 20 recent weddings in a country yielded a mean wedding cost of $26,387.22. Assume that recent wedding costs in this country are normally distributed with a standard deviation of $8400. Complete parts (a) through (c) below. a. Determine a 95% confidence interval for the mean cost, u, of all recent weddings in this country. The 95% confidence interval is from $to $ (Round to the nearest cent as needed.) b. Interpret your result in part (a)....
a
randomnsample of 20 recent weddings in a country yeilded a mean
wedding cost of $26,368.73. assume that recent wedding costs in
this country are normally distributed with a standard deviation of
$8100. complete parts A-C below
This Question: 1 pt 12 of 12 (5 complete) Arandom sample of 20 ecent weddings in a ceuntry yelded a mean wedding oost of $26 3873 Assume that recent wedding costs in this country ae nomally dstributed wth a standand deviation of 500...
Many couples believe that it is getting too expensive to host an "average" wedding in a certain country. According to a wedding cost website, the average cost of a wedding in the country in 2009 was $24,066. Recently, in a random sample of 40 weddings in the country it was found that the average cost of a wedding was $23,224, with a standard deviation of $2,903. On the basis of this, a 95% confidence interval for the mean cost of...
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