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Define the random variable X¯{"x":[[4,22,36],[6,35],[5.647456684472792,6.517307794504744,6.517307794504744,8.257010014568648,9.1268611246006,9.996712234632552,12.606265564728409,13.476116674760362,14.345967784792315,16.08567000485622,16.95552111488817,17.825372224920123,18.695223334952075,20.43492555501598,21.30477666504793,22.174627775079884,23.044478885111836,23.914329995143788,24.78418110517574,25.654032215207693,26.523883325239645,27.393734435271597,27.393734435271597,28.26358554530355,29.1334366553355,29.1334366553355,30.003287765367453,30.873138875399405,31.742989985431358,32.61284109546331]],"y":[[9,59,116],[115,9],[-7.088088802556169,-7.088088802556169,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-8.827791022620074,-8.827791022620074,-8.827791022620074,-9.697642132652026,-9.697642132652026,-9.697642132652026,-9.697642132652026,-9.697642132652026]],"t":[[0,0,0],[0,0],[1648475418436,1648475418678,1648475418692,1648475418709,1648475418727,1648475418743,1648475418760,1648475418776,1648475418791,1648475418810,1648475418826,1648475418842,1648475418859,1648475418875,1648475418892,1648475418910,1648475418925,1648475418942,1648475418965,1648475419009,1648475419026,1648475419076,1648475419105,1648475419118,1648475419134,1648475419146,1648475419161,1648475419180,1648475419242,1648475419292]],"version":"2.0.0"}in words.

Short Answer

Expert verified

X¯represents the sample mean.

Step by step solution

01

Given information

Given in the question that, One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of151hours each month with a standard deviation of 32hours. Assume that the underlying population distribution is normal.

02

Explanation

The average amount of time Americans spend watching television every month.

The statistic X¯is a measure on a sample that is used to calculate the mean of the population from which it is drawn, that is, μ.

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Most popular questions from this chapter

Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be % confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones?

Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90%? Why?

Suppose that 14children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had

to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of

three months. Assume that the underlying population distribution is normal.

a.i.x̄=__________ii.sx=__________iii.n=__________iv.n–1=__________

b. Define the random variable Xin words.

c. Define the random variable Xin words.

d. Which distribution should you use for this problem? Explain your choice.

e. Construct a 99%confidence interval for the population mean length of time using training wheels.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

f. Why would the error bound change if the confidence level were lowered to 90%?

A sample of 16small bags of the same brand of candies was selected. Assume that the population distribution of bag

weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation

of 0.12ounces. The population standard deviation is known to be0.1 ounce.

a.i.x̄=________ii.σ=________iii.sx=________

b. In words, define the random variableX.

c. In words, define the random variableXÌ„.

d. Which distribution should you use for this problem? Explain your choice.

e. Construct a 90%confidence interval for the population mean weight of the candies.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

f. Construct a 98%confidence interval for the population mean weight of the candies.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

g. In complete sentences, explain why the confidence interval in partf is larger than the confidence interval in part e.

h. In complete sentences, give an interpretation of what the interval in part fmeans.

Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample mean.

See all solutions

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