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Paha777 [63]
3 years ago
9

Find the critical value of x2 based on the given information H1 >3.5 n=14 a=0.05

Mathematics
1 answer:
egoroff_w [7]3 years ago
4 0

Answer:

The degrees of freedom are given by:

df = n-1=14-1=13

The significance level is \alpha=0.05 and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

\chi^2_{\alpha}= 22.362

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

Step-by-step explanation:

We have the followign system of hypothesis:

Null hypothesis: \sigma^2 \leq 3.5

Alternative hypothesis: \sigma^2 >3.5

The degrees of freedom are given by:

df = n-1=14-1=13

The significance level is \alpha=0.05 and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

\chi^2_{\alpha}= 22.362

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

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2 years ago
Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams
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Answer:

a) 0.2318

b) 0.2609

c) No it is not unusual for a broiler to weigh more than 1610 grams

Step-by-step explanation:

We solve using z score formula

z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.

(a) What proportion of broilers weigh between 1150 and 1308 grams?

For 1150 grams

z = 1150 - 1387/192

= -1.23438

Probability value from Z-Table:

P(x = 1150) = 0.10853

For 1308 grams

z = 1308 - 1387/192

= -0.41146

Probability value from Z-Table:

P(x = 1308) = 0.34037

Proportion of broilers weigh between 1150 and 1308 grams is:

P(x = 1308) - P(x = 1150)

0.34037 - 0.10853

= 0.23184

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(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?

1510 - 1387/192

= 0.64063

Probabilty value from Z-Table:

P(x<1510) = 0.73912

P(x>1510) = 1 - P(x<1510) = 0.26088

≈ 0.2609

(c) Is it unusual for a broiler to weigh more than 1610 grams?

1610- 1387/192

= 1.16146

Probability value from Z-Table:

P(x<1610) = 0.87727

P(x>1610) = 1 - P(x<1610) = 0.12273

≈ 0.1227

No it is not unusual for a broiler to weigh more than 1610 grams

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Answer:

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