HELP PLEASE (Geometric Sequences) 15 points

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.

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3 years ago

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STatiana [176]

Answer:

275 miles

Step-by-step explanation:

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What is lcm of 54 and 64

ankoles [38]

The lcm is 8

7×8=56

6×6=64

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What is -1 3/5 ÷ -2/3?

elena55 [62]

Answer:

0.26666666666

Step-by-step explanation:

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3 years ago

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The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.

AURORKA [14]

The two nominal variables are related.

The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.

Ownership C1 C2 C3

No 40 32 48

Yes 30 48 52

The first thing to do in order to determine if they are related or not is to state our null and alternative hypothesis

Null hypothesis

$\mathbf{H_o: Two \nomial \ variables \ are \ related}$

Alternative hypothesis

$\mathbf{H_a: Two \nomial \ variables \ are \ not \ related}$

Using the Chi-square test statistics which can be expressed by using the formula

$X ^2 = \sum \dfrac{(O-E)^2}{E}$

Ownership C1 C2 C3 Total

No 40 32 48 120

Yes 30 48 52 130

Total 70 80 100 250

The expected values are calculated as:

$\mathsf{E_{a,b} = \dfrac{(row \ total \times column \ total )}{grand \ total }}$

$\mathsf{E_{1,1} = \dfrac{(70 \times120 )}{250 }}$

$\mathsf{E_{1,1} = 33.6}$

$\mathsf{E_{1,2} = \dfrac{(70 \times130 )}{250 }}$

$\mathsf{E_{1,2} = 36.4}$

$\mathsf{E_{2,1} = \dfrac{(80 \times120 )}{250 }}$

$\mathsf{E_{2,1} = 38.4}$

$\mathsf{E_{2,2} = \dfrac{(80 \times 130 )}{250 }}$

$\mathsf{E_{2,2} = 41.6}$

$\mathsf{E_{3,1} = \dfrac{(100 \times120 )}{250 }}$

$\mathsf{E_{3,1} = 48}$

$\mathsf{E_{3,2} = \dfrac{(70 \times130 )}{250 }}$

$\mathsf{E_{3,2} = 52}$

∴ Using the Chi-square test statistics, we have:

$X ^2 = \sum \dfrac{(O-E)^2}{E}$

$X ^2 = \Bigg( \dfrac{(40-33.6)^2}{33.6}+ \dfrac{(30-36.4)^2}{36.4}+ \dfrac{(32-38.4)^2}{38.4}+ \dfrac{(48-41.6)^2}{41.6} + \dfrac{(48-48)^2}{48}+ \dfrac{(52-52)^2}{52} \Bigg)$

$X ^2 = \Bigg( \dfrac{40.96}{33.6}+ \dfrac{40.96}{36.4}+ \dfrac{40.96}{38.4}+ \dfrac{40.96}{41.6} + \dfrac{0}{48}+ \dfrac{0}{52} \Bigg)$

$X ^2 = \Bigg( 1.2190+ 1.1253+ 1.0667+ 0.9846+0+ 0 \Bigg)$

$\mathbf{X ^2 =4.3956}$

The degree of freedom df = ((r - 1) × (c - 1))

= (3 - 1) (2 -1 )

= 2 × 1

= 2

∴

Assuming the level of significance = 5%

The p-value of the Chi-square test statistics at df of 2 is:

= $\mathbf{P(X^2 > 4.3956) \implies 0.111}$

Therefore, we can conclude that since the p-value (0.111) is greater than the level of significance (0.05), we fail to reject the null hypothesis.

Hence, the two nominal variables are related.

Learn more about Chi-square test statistics here:

brainly.com/question/2365682?referrer=searchResults

7 0

2 years ago