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Solution — 2025 Question 12 Sampling| Hypothesis Testing

(a)
Unbiased estimate of population mean=x‾=∑xn=281680=1765=35.2
Unbiased estimate of population variance=s2=1n−1(∑x2−(∑x)2n)=180−1(99187−(2816)280)=319395

Let X denote the random variable of the salinity (in g/kg) of a randomly chosen seawater specimen
Let μ denote the population mean salinity (in g/kg) of a randomly chosen seawater specimen

H0:μ=35
H1:μ>35

Under H0, X‾∼N(35,319395280) Z=X‾−35319395280∼N(0,1) approximately by the Central Limit Theorem since n=80 is large

From GC,
p-value = 0.023264≤0.05⇒Reject H0

Hence there is sufficient evidence at the 5% level of significance to conclude that the mean salinity of seawater in the area has increased ∎

(b)

H0:μ=35
H1:μ≠35

where μ g/kg is the population mean salinity in the different area.

Since there is insufficient evidence to reject H0 at the 5% significance level,

−1.9600<z<1.9600−1.9600<x‾−μsn<1.9600

Since x‾−μ=35.08−35>0,

x‾−μsn<1.960035.08−35s100<1.96000.8s<1.96000.8<1.9600s1.9600s>0.8s>0.40817

Let σx be the standard deviation of the sample

s2=nn−1σx2=100100−1σx2=10099σx2s=10099σx
10099σx>0.40817σx>0.1224511σx>0.406∎ (to 3 s.f.)
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Parts

  • Part (a)
  • Part (b)