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PSYCHO15rus [73]
2 years ago
5

An epidemiologist found five cases of "big toe cancer" in the Yukon Territory. Because there were only a few cases, the epidemio

logist decided to conduct a matched case-control study to determine whether shoe size larger than 9 is a risk factor for big toe cancer. Cases were individually matched to one control for daily activity, history of athlete’s foot, and history of ingrown toenails. The following data were gathered:
Shoe size > 9
Pair Case Control
1 Yes No
2 No No
3 No Yes
4 Yes Yes
5 No Yes
Compute the proper measure of association.
Interpret your results.
If you were to investigate a rare cancer in Lynchburg, where might you look for data?
What would be necessary legally and ethically to be able to utilize this data set(s)?
Submit your thread by 11:59 p.m. (ET) on Thursday of Module/Week 3, and submit your replies by 11:59 p.m. (ET) on Sunday of the same module/week.
Mathematics
1 answer:
bulgar [2K]2 years ago
6 0

Answer:

Step-by-step explanation:

Given that:

An epidemiologist found five cases of "big toe cancer" in the Yukon Territory.

Therefore, shoe size > 9

1) From the required data given below

                       <u>Case       Control       Total</u>

Yes                2(a)             3(b)            5

No                  3(c)             2(d)            5

Total               5                 5                10

∴ odds ratio = ad/bc

= 4/9

=0.444

2) From the less than 1.0 mean that the odds of cancer among case is lower than the odds of cancer among controls

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12.80+3.42= 16.22

16.22-9.70= 6.52 that's your answer

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If the egress rate for a stadium is 5,500 fans per hour and it takes fans about one-third less time for fans to exit a game as i
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Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The
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Answer:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

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

And the statistic is given by this formula:

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

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

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

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

We have the following data given:

n_1 =20 represent the sample size for group 1

n_2 =20 represent the sample size for group 2

\bar X_1 =43.5 represent the sample mean for the group 1

\bar X_2 =40.1 represent the sample mean for the group 2

s_1=4.1 represent the sample standard deviation for group 1

s_2=3.2 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525

And the deviation would be just the square root of the variance:

S_p=3.678

The statistic is givne by:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0
2 years ago
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