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Harman [31]
2 years ago
15

Diane serves breakfast to two groups of children at a daycare center. One box of oaties contains 12 cups of cereal.She needs 1/3

cup for each younger child and 3/4 cup for each older child. Today's group includes 11 younger children and 10nolder childen.is one box of oaties enough for everyone, explain.
Mathematics
1 answer:
IRISSAK [1]2 years ago
6 0
11(1/3)+(3/4)10 = 11/3+30/4
44/12+60/12=104/4 = 26 cups one box is not enough u will need three 
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sergejj [24]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958      

c) p_v =P(t_{34}>2.958)=0.0028    

d) If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

The sample mean is given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And the sample deviation is:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=17 represent the sample mean    

s=4 represent the sample standard deviation

n=35 sample size  

\mu_o =15 represent the value that we want to test  

\alpha=0.01 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 check if the mean is less or equal than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

Since we don't  know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-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".  

Part b Calculate the statistic  

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

t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958  

Part c P-value  

The degrees of freedom are given by:

df = n-1= 35-1=34

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

p_v =P(t_{34}>2.958)=0.0028  

Part d Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.  

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