Distance= 2797.2 time= 0.15 speed

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago

Distributions and Comparing Data Project

LenKa [72]

Answer:

1. The shape of the histogram created with MS Excel is approximately bell shaped and approximately evenly spread about the central (largest counts) values

The shape of the box plot created with MS Excel is approximately evenly distributed

2. The appropriate measure of central tendency for a bell shaped histogram is the mean and median, due to the approximately equal distribution about the highest frequency class

3. Part A

The 15 numbers between 1 and 20 generated by a random generator are;

1, 8, 8, 15, 4, 18, 11, 6, 17, 1, 18, 15, 10, 12, 11

The measure of center is the mean = (1+8+8+15+4+18+11+6+17+1+18+15+10+12+11)/15 = 155/15 = 31/3

The measure of spread used is the variance, s² = 32.38

$Where, \,s^2 =\dfrac{\sum \left (x_i-\bar x \right )^{2} }{n - 1}$

${\sum \left (x_i-\bar x \right )^{2} }$ = 453.333

n = 15

s² = 453. $\overline 3$ /(15 - 1) = 32.38

Part B

The measure of central tendency are the mean and the median because the size of the data (n = 15), is 75% of the population (N = 20) nd therefore the data is approximately normal and can be represented by the mean and the standard deviation

Step-by-step explanation: