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Evgesh-ka [11]
1 year ago
5

HELP 64 POINTS

Mathematics
2 answers:
xxTIMURxx [149]1 year ago
7 0

Answer:

{30, 39, 33, 15, 15, 18, 20, 33, 19, 41, 45, 18}

kodGreya [7K]1 year ago
5 0

Answer:  {30, 39, 33, 15, 15, 18, 20, 33, 19, 41, 45, 18}

Step-by-step explanation:

1. The key indicates you how to read the values of the stem-and-leaf plot. Then, 1|5 means 15.

2. Then, the first digit is written in the colum Stem and the second digit in the column Leaf.

3. Based on this information, you can write the data given in the stem-and-leaf plot, as following:

15, 15, 18, 18, 19 (The first digit is 1 and the second digit are: 5,5,8,8 and 9)

20 (The first digit is 2 and the second digit is 0)

30, 33, 33, 39 (The first digit is 3 and the second digit are: 0,3,3 and 9)

41, 45 (The first digit is 4 and the second digit are: 1 and 5)

3. Then, as you can see, the data set represented by the stem-and-leaf plot is:

{30, 39, 33, 15, 15, 18, 20, 33, 19, 41, 45, 18}

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C

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A newly hired basketball coach promised a high-paced attack that will put more points on the board than the team’s previously te
puteri [66]

Answer:

a. z = 2.00

Step-by-step explanation:

Hello!

The study variable is "Points per game of a high school team"

The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)

The hypothesis is:

H₀: μ ≤ 99

H₁: μ > 99

α: 0.01

There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.

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7 0
1 year ago
The following are the hypotheses for a test of the claim that college graduation statue and cola preference are independent. H0:
Shkiper50 [21]

Answer:

\chi^2 = 0.579

And the critical value for the significance level used is:

\chi^2_{crit}= 5.991

Since the calculated value is less than the critical value we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the College graduation status and cola preference are independent

Step-by-step explanation:

For this case we want to test the following hypothesis:

Null hypothesis: College graduation status and cola preference are independent

Alternative hypothesis: College graduation status and cola preference are dependent

For this case we got a calculated statistic of:

\chi^2 = 0.579

And the critical value for the significance level used is:

\chi^2_{crit}= 5.991

Since the calculated value is less than the critical value we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the College graduation status and cola preference are independent

7 0
1 year ago
A chi-square test for goodness of fit is used to examine the distribution of individuals across three categories, and a chi-squa
aleksandrvk [35]

Answer:

Equal df

Step-by-step explanation:

Given that a chi square test for goodness of fit is used to examine the distribution of individuals across three categories,

Hence degree of freedom = 3-1 =2

Similarly for a chi-square test for independence is used to examine the distribution of individuals in a 2×3 matrix of categories.

Here degree of freedom = (r-1)(c-1) where r = no of rows and c =no of columns

= (2-1)(3-1) = 2

Thus we find both have equal degrees of freedom.

3 0
1 year ago
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