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marissa [1.9K]
3 years ago
10

How to solve 9x 4 2/5 in words

Mathematics
1 answer:
Natalija [7]3 years ago
3 0

Answer:

39.6

Thirty nine point six

Step-by-step explanation:

9 x 4 2/5

9 x 22/5

198/5 = 39.6

Thirty nine point six

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Answer:

z=\frac{1.83-1.94}{\sqrt{\frac{1.48^2}{180}+\frac{1.62^2}{180}}}}=-0.673  

p_v =2*P(z

Comparing the p value with the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the two means

Step-by-step explanation:

Data given and notation

\bar X_{1}=1.83 represent the mean in 1990

\bar X_{2}=1.94 represent the mean for 2005

s_{1}=1.48 represent the sample deviation for 1990

s_{2}=1.62 represent the sample standard deviation for 2005

n_{1}=180 sample size for 1990

n_{2}=180 sample size for 2005

t would represent the statistic (variable of interest)

\alpha=0.05 significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:\mu_{1}=\mu_{2}

Alternative hypothesis:\mu_{1} \neq \mu_{2}

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

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

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.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

z=\frac{1.83-1.94}{\sqrt{\frac{1.48^2}{180}+\frac{1.62^2}{180}}}}=-0.673  

What is the p-value for this hypothesis test?

Since is a bilateral test the p value would be:

p_v =2*P(z

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the two means

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