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AveGali [126]
3 years ago
13

A local florist charges $15 for delivery plus $2.75 per flower for all flower arrangements. When Marcus got the bill for a flowe

r arrangement he ordered, it was less than $50. Which of the following could not represent the number of flowers in the arrangement Marcus ordered?
Mathematics
1 answer:
ikadub [295]3 years ago
7 0
What are the options????
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2,231.5= how many grams?
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Find dy/dx for (y^2/x) = 18 by implicit differentiation
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  dy/dx = y/(2x)

Step-by-step explanation:

The product formula can be used, along with the power rule.

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Under which condition does a light ray striking a concave mirror reflect off parallel to the principal axis?. . . A. when the in
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A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. The research fir
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Answer:

a) F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73

b) For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be \alpha/2 =0.025, the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: F>4.026 \cup F since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Step-by-step explanation:

Data given and notation  

n_1 = 10 represent the sampe size for the Miller's Stores

n_2 =10 represent the sample size for the Albert's stores

\bar X_1 =121.92 represent the sample mean for Miller's store

\bar X_2 =114.81 represent the sample mean for Albert's store

s_1 = 1.4 represent the sample deviation for the Miller's store

s_2 = 1.84 represent the sample deviation for the Albert's stores

s^2_2 = 12.25 represent the sample variance for the utility stocks

\alpha=0.05 represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

F=\frac{s^2_2}{s^2_1}

Solution to the problem  

System of hypothesis

We want to test if the variation for th two groups is the same, so the system of hypothesis are:

H0: \sigma^2_1 = \sigma^2_2

H1: \sigma^2_1 \new \sigma^2_2

a) Calculate the statistic

Now we can calculate the statistic like this:

F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have n_1 -1 =10-1=9 and for the denominator we have n_2 -1 =10-1=9 and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:

P value

p_v =2*P(F_{9,9}>1.727)=0.428

And we can use the following excel code to find the p value:"=2*(1-F.DIST(1.727,9,9,TRUE))"

b) Critical value

For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be \alpha/2 =0.025, the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: F>4.026 \cup F since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Since our calcu

Conclusion

Since the p_v > \alpha we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the two deviations are different at 5% of significance.  

8 0
3 years ago
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