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ArbitrLikvidat [17]
3 years ago
10

Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate

, and state the final conclusion that addresses the original claim. A cereal company claims that the mean weight of the cereal in its packets is 14 oz. The weights (in ounces) of the cereal in a random sample of 8 of its cereal packets are listed below. 14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2 Test the claim at the 0.01 significance level.H0: μ = 14 oz. H1: μ 14 oz. Test statistic: t = 0.408 Critical values: t = ±3.499. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the mean weight is 14 ounces.Assume that you plan to use a significance level of α = 0.05 to test the claim that p1 = p2, Use the given sample sizes and numbers of successes to find the pooled estimate p. n1 = 100 n2 = 100 x1 = 33 x2 = 36 A) 0.241 B) 0.380 C) 0.310 D) 0.345
Mathematics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that  A cereal company claims that the mean weight of the cereal in its packets is 14 oz

Test the claim at the 0.01 significance level.H0: μ = 14 oz. H1: μ 14 oz.

Parameter  Value

Mean  14.050

SD  0.346

SEM  0.122

N  8      

90% CI  13.818 to 14.282

95% CI  13.760 to 14.340

99% CI  13.621 to 14.479

Minimum  13.6

Median  14.05

Maximum  14.6

Test statistic: t = 0.408

Critical values: t = ±3.499.

Since test statistic falls within the critical values, we accept the null hypothesis at 1% significance.

Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the mean weight is 14 ounces.

Assume that you plan to use a significance level of α = 0.05 to test the claim that p1 = p2, Use the given sample sizes and numbers of successes to find the pooled estimate p. n1 = 100 n2 = 100 x1 = 33 x2 = 36

proportion difference = 0.03

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A random sample of 850 births included 434 boys. Use a 0.10 significance level to test the claim that 51.5​% of newborn babies a
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Answer:

A

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z = -0.257

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Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 90% confidence interval) ---- reject Null hypothesis

Z score < Z(at 90% confidence interval) ------ accept Null hypothesis

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

For the case above;

Null hypothesis: H0: p = 0.515

Alternative hypothesis: Ha ≠ 0.515

Given;

n=850 represent the random sample taken

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 850

po = Null hypothesized value = 0.515

p^ = Observed proportion = 434/850 = 0.5106

Substituting the values we have

z = (0.5106-0.515)/√(0.515(1-0.515)/850)

z = −0.256677

z = −0.257

To determine the p value (test statistic) at 0.10 significance level, using a two tailed hypothesis.

P value = P(Z<-0.257) = 0.797

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = -0.257) which falls with the region bounded by Z at 0.10 significance level. And also the one-tailed hypothesis P-value is 0.797 which is greater than 0.10. Then we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that at 10% significance level the null hypothesis is valid.

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