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kkurt [141]
3 years ago
15

Sam and lauren are going on a date. For dinner, their options are thai or italian, then they will either go to a movie, go ice s

kating or go to an art museum. How many different combinations are their for their date?
Mathematics
1 answer:
kolbaska11 [484]3 years ago
6 0

Answer: 6

Step-by-step explanation:

There are 6 different combinations of choices for the date:

  1. Thai food, movie
  2. Thai food, ice skating
  3. Thai food, art museum
  4. Italian food, movie
  5. Italian food, ice skating
  6. Italian food, art museum

An easier way to solve for this is multiply by how many choices are in each category.

In food there is 2 choices, in activities there is 3 choices

Therefore:  2 x 3 = 6

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In doing so, you collect a random sample of 50 salespersons employed by his company, which is thought to be representative of sa
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Answer:

z=\frac{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279  

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And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"

Step-by-step explanation:

Assuming this complete problem: "The CEO of a software company is committed to expanding the proportion of highly qualified women in the organization's staff of salespersons. He believes that the proportion of women in similar sales positions across the country is less than 45%. Hoping to find support for his belief, he directs you to test

H0: p .45 vs H1: p < .45.

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Compute the p-value associated with this test. Place your answer, rounded to 4 decimal places, in the blank. For example, 0.3456 would be a legitimate entry."

1) Data given and notation

n=50 represent the random sample taken

X=18 represent the number of women in the sample selected

\hat p=\frac{18}{50}=0.36 estimated proportion of women in the sample

p_o=0.45 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of women is less than 0.45:  

Null hypothesis:p\geq 0.45  

Alternative hypothesis:p < 0.45  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

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

z=\frac{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z

And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"

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Step-by-step explanation:

I hope that helped

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