You are studying two species of wasp. You are interested in the sterile, worker wasps.

Question

You are studying two species of wasp. You are interested in the sterile, worker wasps. You record the lifespan (in days) of a random sample of wasps from each population.

Population 1: (checksum: 313)

20	20	20	16	14	18	20	20	20	21
19	14	12	13	13	18	18	17

Population 2: (checksum: 308)

13	16	19	12	17	11	19	11	19	18
19	12	17	17	14	13	16	14	15	16

Is there evidence, at a 5% level of significance, that there is a difference in mean lifespan for the worker wasps between these two populations?

a) State the parameter of interest, and verify that the necessary conditions are present in order to carry out the inference procedure.

b) Find the critical value and standard error. It is assumed that you are using the more accurate estimate for the degree of freedom. It is recommended that you use technology to find the degree of freedom!

Degree of freedom: 

Standard Error: 

Standardized Test Statistic: 

P-value: 

c) Should the null hypothesis be rejected? (Enter "yes" or "no".) 

d) Write a conclusion in context

Answer 1

Hi Cynthya,

You could do this one of two ways--pooled t or unpooled t. Given that your instructor wants the more accurate method for degrees of freedom, I am guessing s/he wants unpooled t, so we will go with that. I do not have technology available to compute degrees of freedom, though, so we have to do it by hand. Warning in advance: it's a tedious calculation. Anyhow:

(a) Parameter of interest is true population difference in mean lifespan between the two wasp species. We can reasonably assume wasps were randomly selected and no one wasp selection influenced another wasp selection, so we are safe with assumptions of randomness and independence.

(b) Use a calculator--manual or online--to get the means and standard deviations for the two samples:

xbar₁=17.389

s₁=2.973

xbar₂=15.4

s₂=2.761

To get a critical value, we first need degrees of freedom. This is where things get rough. If we don't pool, we have to use the Satterthwaite approximation to degrees of freedom. This is:

df=[(s₁²/n₁) + (s₂²/n₂)]² / [[[(1/(n₁-1))(s₁²/n₁)²] +[[(1/(n₂-1))(s₂²/n₂)²]]]

where:

s₁=first sample standard deviation

s₂=second sample standard deviation

n₁=first sample size

n₂=second sample size

Now, we substitute in:

s₁=2.973

s₂=2.761

n₁=20

n₂=20

This gives:

df=[(2.973²/20) + (2.761²/20)]² / [[[(1/(20-1))(2.973²/20)²] +[[(1/(20-1))(2.761²/20)²]]]

df=37.789

I do not have access to technology that computes critical value directly, so we have to use a t-table for this. The question specified 5% level of significance, two-sided alternative, so we look for 95% column on t-table. For degrees of freedom row, we need to round down for safety, so we look for 37: closest value is 30,

so

tcrit=2.042

Statistical software will get you a more accurate critical value than this, but I do not currently have access to it.

Now, for unpooled t, the t-test statistic formula is:

t=(xbar₁-xbar₂)-D/sqrt[(s₁²/n₁) + (s₂²/n₂)]

Formula for standard error is the denominator:

SE=sqrt[(s₁²/n₁) + (s₂²/n₂)]

Again, we substitute the above values in:

SE=sqrt[(2.973²/20) + (2.761²/20)]

SE=0.907

Now, for the t test statistic:

t=(xbar₁-xbar₂)-D/sqrt[(s₁²/n₁) + (s₂²/n₂)]

xbar₁=17.389

xbar₂=15.4

s₁=2.973

s₂=2.761

n₁=20

n₂=20

D=0, hypothesized value for difference in means

Recall that we already computed the denominator:

t=[(17.389-15.4)-0]/0.907

t=2.192

Now, we can get the p-value from Excel:

p=T.DIST.2T(ABS(t),df)

Note: Use the most precise value for df here.

p=T.DIST.2T(ABS(2.192),37.789)

p=0.035

(c) p<alpha, 0.035 < 0.05, so, yes.

(d) Reject H₀ and conclude some significant difference in true mean lifespan between the two wasp species.

I hope this helps.