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Tpy6a [65]
2 years ago
8

Two random samples were done to determine how often people in a community go to the dentist. the first sample surveyed 10 people

as they entered their workplaces. the first sample found that the mean time between visits to the dentist was 7.5 months. the second sample surveyed 50 people as they entered the local grocery store. the second sample found that the mean time between visits to the dentist was 18.1 months. which statements are true? check all that apply. the first sample is likely to be more representative of the population. the second sample is likely to be more representative of the population. the second sample will give a better representation because it is larger. the first sample is not biased. community members probably visit the dentist about once every 18 months.
Mathematics
1 answer:
uysha [10]2 years ago
5 0
The answer of all of this is 18 because 18.1 months so if you do the math you get 18-7x+20^2
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If A is an obtuse angle and cos A = 3/5, find the values of tan A
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Complete a b and c if you do one it would mean alot
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The FDA regulates that a fish that is consumed is allowed to contain at most 1 mg/kg of mercury. In Florida, bass fish were coll
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Yes. At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.

The random variable is the sample mean amount of mercury in the bass fish from the lakes of Florida.

The population parameter is the mean amount of mercury in the bass fish of Florida lakes.

The alternative hypothesis (Ha) states that the amount of mercury significantly differs from 1 mg/kg.

The null hypothesis (H0) states that the amount of mercury is not significantly different from 1 mg/kg.

H_0: \mu=1\\\\H_a:\mu\neq 1

Step-by-step explanation:

<em>The question is incomplete.</em>

<em>There is no data provided.</em>

<em>We will work with a sample mean of 0.95 mg/kg and sample standard deviation of 0.15 mg/kg to show the procedure.</em>

<em />

This is a hypothesis test for the population mean.

The claim is that the fish in all Florida lakes have different mercury than the allowable amount (1 mg of mercury per kg of fish).

Then, the null and alternative hypothesis are:

H_0: \mu=1\\\\H_a:\mu\neq 1

The significance level is assumed to be 0.05.

The sample has a size n=53.

The sample mean is M=0.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.15.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.15}{\sqrt{53}}=0.0206

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{0.95-1}{0.0206}=\dfrac{-0.05}{0.0206}=-2.427

The degrees of freedom for this sample size are:

df=n-1=53-1=52

This test is a two-tailed test, with 52 degrees of freedom and t=-2.427, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t

As the P-value (0.019) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.

8 0
4 years ago
HELP ASAP!!
BaLLatris [955]
I would say A. The volume of prism z is 4 times the prims y, and they are similar.

4 x6 is 24
They are similar
That’s really all you need to know.
Hope this helps!
- Pam Pam
4 0
3 years ago
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