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A random sample of 64 bags of white cheddar popcorn are weighed. On average, they weight 5.23 ounces with a standard deviation of 0.24 ounce. Test the hypothesis that = 5.5 ounces against the alternative hypothesis, <5.5 ounces, at the 0.05 level of significance.

In the above, it's not clear to me what you mean by = 5.5.

If you mean "the true mean of the production line is 5.5", I would assume that bag weights are normally distributed. So I would calculate the t-statistics given the sample of n=64, mean=5.23, and sd=0.24, in R:

diff <- 5.23 - 5.5
stderr <- (0.24/sqrt(64))
tstat <- diff/stderr # => -9

How extreme is this t-statistics?

p <- pt(tstat, df=63)

p ~ 3.24e-13 which means that in a world where the factory produces with a mean of 5.5, you are extremely unlikely to see a sample of n=64 with mean as low as 5.23 and sd=0.24.

If instead you mean "a bag plucked at random from the production line is < 5.5 ounces". I would calculate the z-score given the observed mean, standard deviation and reference weight (5.5):

z <- (5.23 - 5.5)/0.24

Than map this value to a t-distribution with 64-1 degrees of freedom:

pt(z, df=63)
[1] 0.132

This means that given the data and assumptions, ~13% of the bags is expected to weigh 5.5 ounces or more.

Either way, I don't see the point in setting a significance threshold, at least in the way the question is phrased.

A random sample of 64 bags of white cheddar popcorn are weighed. On average, they weight 5.23 ounces with a standard deviation of 0.24 ounce. Test the hypothesis that = 5.5 ounces against the alternative hypothesis, <5.5 ounces, at the 0.05 level of significance.

In the above, it's not clear to me what you mean by = 5.5.

If you mean "the true mean of the production line is 5.5", I would assume that bag weights are normally distributed and independent from each other (they are i.i.d). So I would calculate the t-statistics given the sample of n=64, mean=5.23, and sd=0.24, in R:

diff <- 5.23 - 5.5
stderr <- (0.24/sqrt(64))
tstat <- diff/stderr # => -9

How extreme is this t-statistics?

p <- pt(tstat, df=63)

p ~ 3.24e-13 which means that in a world where the factory produces with weights from normal distribution with mean of 5.5, you are extremely unlikely to see a sample of n=64 with mean as low as 5.23 and sd=0.24.

If instead you mean "a bag plucked at random from the production line is < 5.5 ounces". I would calculate the z-score given the observed mean, standard deviation and reference weight (5.5):

z <- (5.23 - 5.5)/0.24

Then map this value to a t-distribution with 64-1 degrees of freedom:

pt(z, df=63)
[1] 0.132

This means that given the data and assumptions, ~13% of the bags is expected to weigh 5.5 ounces or more.

Either way, I don't see the point in setting a significance threshold, at least in the way the question is phrased.

A random sample of 64 bags of white cheddar popcorn are weighed. On average, they weight 5.23 ounces with a standard deviation of 0.24 ounce. Test the hypothesis that = 5.5 ounces against the alternative hypothesis, <5.5 ounces, at the 0.05 level of significance.

In the above, it's not clear to me what you mean by = 5.5.

If you mean "the true mean of the production line is 5.5", I would assume that bag weights are normally distributed. So I would calculate the t-statistics given the sample of n=64, mean=5.23, and sd=0.24, in R:

diff <- 5.23 - 5.5
stderr <- (0.24/sqrt(64))
tstat <- diff/stderr
p <- pt(tstat, df=63)

p ~ 3.24e-13 which means that in a world where the factory produces with a mean 5.5, you are extremely unlikely to see a sample of n=64 with mean 5.23 and sd=0.24.

If instead you mean "a bag plucked at random from the production line is < 5.5 ounces". I would calculate the z-score given the observed mean, standard deviation and reference weight (5.5):

z <- (5.23 - 5.5)/0.24

Than map this value to a t-distribution with 64-1 degrees of freedom:

pt(z, df=63)
[1] 0.132

This means that given the data and assumptions, ~13% of the bags is expected to weigh 5.5 ounces or more.

Either way, I don't see the point in setting a significance threshold, at least in the way the question is phrased.

A random sample of 64 bags of white cheddar popcorn are weighed. On average, they weight 5.23 ounces with a standard deviation of 0.24 ounce. Test the hypothesis that = 5.5 ounces against the alternative hypothesis, <5.5 ounces, at the 0.05 level of significance.

In the above, it's not clear to me what you mean by = 5.5.

If you mean "the true mean of the production line is 5.5", I would assume that bag weights are normally distributed. So I would calculate the t-statistics given the sample of n=64, mean=5.23, and sd=0.24, in R:

diff <- 5.23 - 5.5
stderr <- (0.24/sqrt(64))
tstat <- diff/stderr # => -9

How extreme is this t-statistics?

p <- pt(tstat, df=63)

p ~ 3.24e-13 which means that in a world where the factory produces with a mean of 5.5, you are extremely unlikely to see a sample of n=64 with mean as low as 5.23 and sd=0.24.

If instead you mean "a bag plucked at random from the production line is < 5.5 ounces". I would calculate the z-score given the observed mean, standard deviation and reference weight (5.5):

z <- (5.23 - 5.5)/0.24

Than map this value to a t-distribution with 64-1 degrees of freedom:

pt(z, df=63)
[1] 0.132

This means that given the data and assumptions, ~13% of the bags is expected to weigh 5.5 ounces or more.

Either way, I don't see the point in setting a significance threshold, at least in the way the question is phrased.

A random sample of 64 bags of white cheddar popcorn are weighed. On average, they weight 5.23 ounces with a standard deviation of 0.24 ounce. Test the hypothesis that = 5.5 ounces against the alternative hypothesis, <5.5 ounces, at the 0.05 level of significance.

In the above, it's not clear to me what you mean by = 5.5.

If you mean "the true mean of the production line is 5.5", I would assume that bag weights are normally distributed. So I would calculate the t-statistics given the sample of n=64, mean=5.23, and sd=0.24, in R:

diff <- 5.23 - 5.5
stderr <- (0.24/sqrt(64))
tstat <- diff/stderr
p <- pt(tstat, df=63)

p ~ 3.24e-13 which means that in a world where the factory produces with a mean 5.5, you are extremely unlikely to see a sample of n=64 with mean 5.23 and sd=0.24.

If instead you mean "a bag plucked at random from the production line is < 5.5 ounces". I would calculate the z-score given the observed mean, standard deviation and reference weight (5.5):

z <- (5.23 - 5.5)/0.24

Than map this value to a t-distribution with 64 degrees of freedom:

pt(z, df=63)
[1] 0.132

This means that given the data, ~13% of the bags weigh 5.5 ounces or more.

Either way, I don't see the point in setting a significance threshold, at least in the way the question is phrased.

A random sample of 64 bags of white cheddar popcorn are weighed. On average, they weight 5.23 ounces with a standard deviation of 0.24 ounce. Test the hypothesis that = 5.5 ounces against the alternative hypothesis, <5.5 ounces, at the 0.05 level of significance.

In the above, it's not clear to me what you mean by = 5.5.

If you mean "the true mean of the production line is 5.5", I would assume that bag weights are normally distributed. So I would calculate the t-statistics given the sample of n=64, mean=5.23, and sd=0.24, in R:

diff <- 5.23 - 5.5
stderr <- (0.24/sqrt(64))
tstat <- diff/stderr
p <- pt(tstat, df=63)

p ~ 3.24e-13 which means that in a world where the factory produces with a mean 5.5, you are extremely unlikely to see a sample of n=64 with mean 5.23 and sd=0.24.

If instead you mean "a bag plucked at random from the production line is < 5.5 ounces". I would calculate the z-score given the observed mean, standard deviation and reference weight (5.5):

z <- (5.23 - 5.5)/0.24

Than map this value to a t-distribution with 64-1 degrees of freedom:

pt(z, df=63)
[1] 0.132

This means that given the data and assumptions, ~13% of the bags is expected to weigh 5.5 ounces or more.

Either way, I don't see the point in setting a significance threshold, at least in the way the question is phrased.