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Lady bird [3.3K]

2 years ago

15

Scenario: Does the age at which babies learn to crawl depend on the time of the year that the babies were born? Data were collec

ted from parents who brought their babies into the University of Denver’s Infant Study Center to participate in one of a number of experiments between 1988 and 1991. Parents reported the birth month and the age in which their child first began to crawl. The resulting data were grouped by month of birth: January, May, and September.

1 answer:

sweet-ann [11.9K]2 years ago

4 0

The age for when babies learn to crawl is different for Januar,May, and September.. so I guess yes?

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Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The filling processes can be assumed to be n

CaHeK987 [17]

Answer:

a) Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

b) $z=\frac{(16.015-16.005)-0}{\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}}=1.350$

c) Since is a bilateral test the p value would be:

$p_v =2*P(z>1.350)=0.177$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different.

d) $0.01-1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=-0.0045$

$0.01+1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=0.025$

So on this case the 95% confidence interval would be given by $-0.0045 \leq \mu_1 -\mu_2 \leq 0.025[/tex] Step-by-step explanation:For this case we have the following info:Machine 1: 16.03 16.01 16.04 15.96 16.05 15.98 16.05 16.02 16.02 15.99Machine 2: 16.02 16.03 15.97 16.04 15.96 16.02 16.01 16.01 15.99 16.00We can calculate the sample mean with the following formula:[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}$

$\bar X_{1}=16.015$ represent the mean for sample 1

$\bar X_{2}=16.005$ represent the mean for sample 2

$\sigma_{1}=0.015$ represent the population standard deviation for 1

$\sigma_{2}=0.018$ represent the sample standard deviation for 2

$n_{1}=10$ sample size for the group 2

$n_{2}=10$ sample size for the group 2

Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

Part a

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

We have the population standard deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

Part b

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$z=\frac{(16.015-16.005)-0}{\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}}=1.350$

Part c

P value

Since is a bilateral test the p value would be:

$p_v =2*P(z>1.350)=0.177$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different.

Part d

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =16.015-16.005=0.01$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Confidence interval

Now we have everything in order to replace into formula (1):

$0.01-1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=-0.0045$

$0.01+1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=0.025$

So on this case the 95% confidence interval would be given by $-0.0045 \leq \mu_1 -\mu_2 \leq 0.025$

5 0

3 years ago

Solve. <br><br> −5.1p + 3.8 = 86.93

Bess [88]

Answer: p = -16.3

Step-by-step explanation:

86.93 - 3.8 = 83.13

83.13/-5.1 = -16.3

5 0

3 years ago

Lisa has a gym membership at Lifetime Fitness. It costs $29.95 to join and she also pays a monthly charge. Lisa has been a membe

RoseWind [281]

Answer:

298.90 - 29.95 = (ANSWER) divide it by 11 then you will have the monthly cost

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The local newspaper has letters to the editor from 20 people. If this number represents 8% of all of the newspaper's readers,

zlopas [31]

let the all readers be x

8%=20×100%÷x

8%x=2000%

x=2000%÷8%

x=250

3 0

3 years ago

Factor the expression 42a+14b using the gcf<br> Please figure this out quickly

valentina_108 [34]

If you are trying to factor it is 14(3a+b).

But if you are looking for the greatest common factor it is 14.

4 0

3 years ago