Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 22, 18, 1
4, 25, 17, 28, 15, 21. Use Table 2.a. Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to 4 decimal places, "sample mean" to 3 decimal places and "sample standard deviation" to 2 decimal places.) Sample mean Sample standard deviation b. Construct the 80% confidence interval for the population mean. (Round "t" value to 3 decimal places, and final answers to 2 decimal places.) Confidence interval to c. Construct the 90% confidence interval for the population mean. (Round "t" value to 3 decimal places, and final answers to 2 decimal places.) Confidence interval to d. What happens to the margin of error as the confidence level increases from 80% to 90%?As the confidence level increases, the interval becomes narrower.As the confidence level increases, the interval becomes wider.
1 answer:
Answer:
a: Sample mean: 20, Sample standard deviation: 1.732
b: 19.13 <µ < 20.87
c: 18.84< µ < 21.16
d: as the confidence level increases, the interval becomes wider
Step-by-step explanation:
a: Sample mean is found by adding up all the individual values of the sample, then dividing by the total number of values in the sample.
Sample standard deviation is the square root of the variance
See the first attached photo for the calculations of these values.
We need to create a 80% confidence interval for the population. Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.
Look for the column on the t-distribution chart that has "area in two tails" of 0.20 (80%), and row 7 (degree of freedom)
The t-value is 1.415
See the second attached photo for the construction of the confidence interval
We need to create a 90% confidence interval for the population. Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.
Look for the column on the t-distribution chart that has "area in two tails" of 0.10 (90%), and row 7 (degree of freedom)
The t-value is 1.895
See the third attached photo for the construction of the confidence interval
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Answer:
The number of unsold cakes was 2
Step-by-step explanation:
<u><em>The question in English is</em></u>
In the school Francisco I. Madero of Ciudad Delicias, the celebration was held for commemorate the arrival of spring, after the parade the stalls were set up of the kermesse. The first grade group bought 8 cakes and sold 3/4 of the total.
How much of the cake was not sold?
Let
x ----> number of cakes sold
y ----> number of cakes that didn't sell
we know that
The first grade group bought 8 cakes
so
-----> equation A
The first grade group sold 3/4 of the total.
so
---> equation B
substitute equation A in equation B
Find the value of y
therefore
The number of unsold cakes was 2
Answer:
A 95% confidence interval for this population proportion is [0.081, 0.159].
Step-by-step explanation:
We are given that a market research company conducted a survey to find the level of affluence in a city.
Out of 267 persons who replied to their survey, 32 are considered affluent.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of people who are considered affluent =
= 0.12
n = sample of persons = 267
p = population proportion
<em>Here for constructing a 95% confidence interval we have used One-sample z-test for proportions.</em>
<u>So, 95% confidence interval for the population proportion, p is ;</u>
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 < < 1.96) = 0.95
P( <
<
) = 0.95
P( < p <
) = 0.95
<u>95% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.081, 0.159]
Therefore, a 95% confidence interval for this population proportion is [0.081, 0.159].