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The population of a city on April 15, 1915, was 47,175. During the period between March 1 and July 1, 1915, 1,325 new cases of s

Norma-Jean [14]

Answer:

the incidence rate is 28.47

Step-by-step explanation:

The computation of the monthly incident rate of active cases is shown below:

Incidence rate is

= 1325 ÷ (47175 - 642) × 1000

= (1325 ÷ 46533 ) × 1000

= 28.47

hence, the incidence rate is 28.47

7 0

3 years ago

Examine the equation below. Use it to answer question 54.

sashaice [31]

D= it's is true because ox equals 3*2x

8 0

3 years ago

Which three-dimensional object is formed when the shape is rotated about the axis as shown?

never [62]

C the sphere i think

7 0

4 years ago

The tensile strength of a fiber used in manufacturing cloth is of interest to the purchaser. Previous experience indicates that

yaroslaw [1]

Answer:

a) $z=\frac{127-125}{\frac{2}{\sqrt{8}}}=2.828$

Since is a one-side upper test the p value would be:

$p_v =P(z>2.828)=1-P(z$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and we can conclude that the true mean is higher than 125

b) We select a one sided interval since the interest is in analyze if the true mean is higher than 125, and then the alternative hypothesis needs to be the alternative hypothesis.

Step-by-step explanation:

Data given and notation

$\bar X=127$ represent the mean for the sample

$\sigma=2$ represent the standard deviation for the population

$n=8$ sample size

$\mu_o =125$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mean is higher than 125:

Null hypothesis: $\mu = 125$

Alternative hypothesis: $\mu > 125$

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$z=\frac{127-125}{\frac{2}{\sqrt{8}}}=2.828$

Calculate the P-value

Since is a one-side upper test the p value would be:

$p_v =P(z>2.828)=1-P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and we can conclude that the true mean is higher than 125

Part b

We select a one sided interval since the interest is in analyze if the true mean is higher than 125, and then the alternative hypothesis needs to be the alternative hypothesis.

5 0

3 years ago

in 2019 there were 16 named storms, of which 8 grew into hurricanes and 5 were major. What fraction of hurricanes were major?

olga2289 [7]

There were 16 total storms, meaning our denominator is 16. Since 8 grew into hurricanes, we can leave that alone. Since 5 out of 16 storms were major, 5 is our numerator.

5/16 storms were major because 8 were hurricanes and the other 3 were other kinds of storms. As a decimal conversion, about 31% of these storms were major.

How to find the percentage conversion:

Step 1: We take the fraction 5/16.

Step 2: Since our numerator is 5 and the default denominator is a maximum of 100, we can multiply 5 by 100.

5 x 100 = 500

Step 3: We can divide 100 by 16 to get our decimal conversion.

100 divided by 16 will give us 31.25, which also is representative of 31 and 1/4 added together.

Your final answer: With there being 16 storms, 5 of these storms are major (5/16). In terms of if there were 100 total storms, about 3/10 of these storms were major.

6 0

2 years ago