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Viefleur [7K]
3 years ago
12

You measure 22 textbooks' weights, and find they have a mean weight of 64 ounces. Assume the population standard deviation is 5.

1 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.
Give your answers as decimals, to two places

Mathematics
1 answer:
lara31 [8.8K]3 years ago
3 0

Answer:

Step-by-step explanation:

We want to determine a 90% confidence interval for the true population mean textbook weight.

Number of sample, n = 22

Mean, u = 64 ounces

Standard deviation, s = 5.1 ounces

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

64 ± 1.645 × 5.1/√22

= 64 ± 1.645 × 1.087

= 64 ± 1.788

The lower end of the confidence interval is 64 - 1.788 = 62.21 ounces

The upper end of the confidence interval is 64 + 1.788 = 65.79 ounces

Therefore, with 90% confidence interval, the true population mean textbook weight is between 62.21 ounces and 65.79 ounces

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Klio2033 [76]
I think the answer is 9...............
8 0
2 years ago
What is the value of 9 in the number below? 2,154.359
melomori [17]

Answer:

9 thousandths

Step-by-step explanation:

3 is tenths

5 is hundredths

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7 0
3 years ago
Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below.
garik1379 [7]

Answer:

The 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =9.2 represent the sample mean 1

\bar X_2 =8.8 represent the sample mean 2

n1=27 represent the sample 1 size  

n2=30 represent the sample 2 size  

\sigma_1 =0.3 population standard deviation for sample 1

\sigma_2 =0.1 population standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =9.2-8.8=0.4

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that t_{\alpha/2}=1.96  

Now we have everything in order to replace into formula (1):  

0.4-1.96\sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}=0.281  

0.4+1.96\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=0.519  

So on this case the 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

6 0
3 years ago
The length of a rectangle is 4 cm longer than its width. If the perimeter of the rectangle is 44 cm, find its area.​
adell [148]
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Width:w
Perimeter:44
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P=44
44=2(w)+2(w+4)
44=2w+2w+8
44=4w+8
36=4w
w=9
L=w+4
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6 0
2 years ago
Read 2 more answers
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