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TEA [102]
3 years ago
15

The null and alternative hypotheses are given. Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tail

ed. What parameter is being​ tested?
Upper H0​: sigmaσ equals= 99
Upper H1​: sigmaσ not equals≠ 99
What type of test is being conducted in this​ problem?
Mathematics
1 answer:
olganol [36]3 years ago
7 0

Answer:

Two tailed test

Parameter tested= population standard deviation

Step-by-step explanation:

1) Important concepts

The chi-square test if the standard deviation of a population is equal to a given value. We can conduct the test two-side or a one-side.

For the case of two-side we want to check if the population deviation is equal to some value or no. And for the case of one-side we want to check if the population devition is higher or lower than a given value.

The system of hypothesis are:

\sigma=\sigma_o Null hypothesis

\sigma \neq \sigma_o Alternative hypothesis

That's when we use the two sided version but we can have other's alternative hypothesis if we use one-side test:

\sigma < \sigma_o Alternative hypothesis

\sigma > \sigma_o  Alternative hypothesis

The test statistic to check the hypothesis is:

t=(N-1)(\frac{s}{\sigma_o})^2

For this special case the value of \sigma_o =99

Where N represent the sample size and s is the sample standard deviation. The ratio \frac{s}{\sigma_o} is important since allows to compare the ratio of the sample standard deviation to the specified value. If this ratio is different from 1, we will have evidence that we can reject the null hypothesis.

Using a significance level \alpha we can find the critical region for the possible alternative hypothesis:

t>\chi_{\alpha,N-1}^2 for the upper one tailed option

t for the lower one tailed option

t for the two tailed option

The N-1 represent the degrees of freedom for the chi square distribution.

2) Solving the question

Based on all the info from above we can conclude that:

Type of test= Two tailed test

Parameter tested=population Standard deviation

And they want to check if the population deviation differs from 99

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Each of 16 students measured the circumference of a tennis ball by four different methods, which were:Method A: Estimate the cir
lys-0071 [83]

Answer:

a) \bar X_A =22.744

\bar X_B =20.7

\bar X_C =21.013

\bar X_D =18.306

b) Median_A =\frac{23+24}{2}=23.5

Median_B =\frac{20.4+20.4}{2}=20.4

Median_C =\frac{21+21}{2}=21

Median_D =\frac{20.7+20.7}{2}=20.7

c) \bar X_A =23.25

\bar X_B =20.7

\bar X_C =21.04

\bar X_D =20.69

Step-by-step explanation:

Data given

Assuming the following data:

Method A: 18.0, 18.0, 18.0, 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0,25.0,26.0,26.4

Method B: 18.8, 18.9, 18.9, 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0, 22.0,22.0,23.6

Method C: 20.2, 20.5, 20.5, 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5,21.5,21.5,21.6

Method D: 20.0, 20.0, 20.0,-20.0, 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5, 21.6, 22.1,22.3.

Part a

The sample mean is defined as:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

If we apply this formula for the four methods we got:

\bar X_A =22.744

\bar X_B =20.7

\bar X_C =21.013

\bar X_D =18.306

Part b

Since the total number of points for each method is 16 and this is an even number we need to calculate the median as the average between the 8th and the 9th position of the data ordered from the smallest to the largest. If we do this we have that:

Median_A =\frac{23+24}{2}=23.5

Median_B =\frac{20.4+20.4}{2}=20.4

Median_C =\frac{21+21}{2}=21

Median_D =\frac{20.7+20.7}{2}=20.7

Part c

The trimmed mean by 20% means that we need to calculate removing the 20% from each of the tails, the 20% of 16 is 3.2, so we need to remove 3 observations from both ends of the data like this:

Method A: 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0

Method B: 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0

Method C: 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5

Method D: 20.0 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5.

If we calculate the mean for each method we got:

\bar X_A =23.25

\bar X_B =20.7

\bar X_C =21.04

\bar X_D =20.69

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