Jeff had $20. He spent

A sport statistician gathered data on the number of points scored in each game by high school basketball players across the coun

Sonbull [250]

For a sample size of 20 players, 68% of the sample means fall within 19.32 and 20.98

<h3>What is a sample size?</h3>

The sample size is a term used in market research for defining the number of subjects included in a sample size. By sample size, we understand a group of subjects that are selected from the general population and is considered a representative of the real population for that specific study.

Empirical rule states that for a normal distribution, 68% of the values are within one standard deviation from the mean, 95% of the values are within two standard deviation from the mean and 99.7% of the values are within three standard deviation from the mean.

For a population mean of 20.15 and a standard deviation of 3.7

68% are within μ ± σ/√n,

hence,

68% = 20.15 ± 3.7/√20 = (19.32, 20.98)

For a sample size of 20 players, 68% of the sample means fall within 19.32 and 20.98.

To learn more about sample size from the given link:

brainly.com/question/22927560

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6 0

2 years ago

Someone helpp !! I don’t understand this at all

Naddika [18.5K]

Answer:

∠AB=9

Step-by-step explanation:

the pyagrothyes therom= $a^{2}$ + $b^{2}$ = $c^{2}$

to find ∠AB we need to find the height of the triangle.

replace the variables:

$a^{2}$ + $6^{2}$ = $9^{2}$

= $9^{2}$ - $6^{2}$ = $a^{2}$

simplify:

81-36= $\sqrt{45}$

now we solve for ∠AB

replace the variables:

$6^{2}$ + $\sqrt{45}$ = $c^{2}$

simplify:

36+45= $\sqrt{81}$

$\sqrt{81}$ =9

I hope this helps you!!

have a nice day!!

4 0

2 years ago

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men Women

The sample mean : 24.51 22.69

Sample standard deviation : 4.48 3.86

Sample size : 35 40

Let $\mu_1$ = mean number of times men order take-out dinners in a month.

$\mu_2$ = mean number of times women order take-out dinners in a month

(a) So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq$ 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_-_n_2_-_2$

where, $\bar X_1$ = sample mean for men = 24.51

$\bar X_2$ = sample mean for women = 22.69

$s_1$ = sample standard deviation for men = 4.48

$s_2$ = sample standard deviation for women = 3.86

$n_1$ = sample of men = 35

$n_2$ = sample of women = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }$ = 4.16

So, test statistics = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

P-value = P( $t_7_3$ > 1.89) = 0.0331

So, P-value for two tailed test is = 2 $\times$ 0.0331 = 0.0662